GQB_EE.pdf

GATE QUESTION BANK for

Electrical Engineering By

GATE QUESTION BANK

Contents

Contents #1.

#2.

Subject Name Mathematics

Topic Name

Page No. 1-148

1 2 3 4 5 6 7

Linear Algebra Probability & Distribution Numerical Methods Calculus Differential Equations Complex Variables Laplace Transform

1 – 28 29 – 57 58 – 73 74 – 112 113 – 131 132 – 143 144 – 148

Network Theory 8 9 10 11 12 13

#3.

19

168 – 185 186 – 203 204 – 206 207 – 214 215 – 216

217 – 223 224 – 238 239 – 250 251 – 256 257 – 261 262 – 275

276 – 340 Basics of Control System Time Domain Analysis Stability & Routh Hurwitz Criterion Root Locus Technique Frequency Response Analysis using Nyquist plot Frequency Response Analysis using Bode Plot Compensators & Controllers State Variable Analysis

Analog Circuits 28 29 30 31 32 33 34

149 – 167

217 – 275 Introduction to Signals & Systems Linear Time Invariant (LTI) systems Fourier Representation of Signals Z-Transform Laplace Transform Frequency response of LTI systems and Diversified Topics

Control Systems 20 21 22 23 24 25 26 27

#5.

Network Solution Methodology Transient/Steady State Analysis of RLC Circuits to DC Input Sinusoidal Steady State Analysis Laplace Transform Two Port Networks Network Topology

Signals & systems 14 15 16 17 18

#4.

149 – 216

276 – 282 283 – 294 295 – 300 301 – 308 309 – 316 317 – 322 323 – 329 330 – 340

341 – 421 Diode Circuits - Analysis and Application AC & DC Biasing-BJT and FET Small Signal Modeling Of BJT and FET BJT and JFET Frequency Response Feedback and Oscillator Circuits Operational Amplifiers and Its Applications Power Amplifiers th

th

341 – 353 354 – 363 364 – 372 373 – 375 376 – 381 382 – 420 421 th

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GATE QUESTION BANK

#6.

Digital Circuits 35 36 37 38 39 40 41 42

#7.

422 – 424 425 – 430 431 – 435 436 – 438 439 – 456 457 – 462 463 – 464 465 – 472

473 – 483 Electromagnetic Field

473 – 483

Measurement 44 45 46 47 48

#9.

422 – 472 Number Systems & Code Conversions Boolean Algebra & Karnaugh Maps Logic Gates Logic Gate Families Combinational and Sequential Digital Circuits AD/DA Convertor Semiconductor Memory Introduction to Microprocessors

EMT 43

#8.

Contents

484 – 516 Basics of Measurements and Error Analysis Measurements of Basic Electrical Quantities 1 Measurements of Basic Electrical Quantities 2 Electronic Measuring Instruments 1 Electronic Measuring Instruments 2

Power Systems 49 50 51 52 53 54

517 – 553 Transmission and Distribution Economics of Power Generation Symmetrical Components & Faults Calculations Power System Stability Protection & Circuit Breakers Generating Stations

#10. Power Electronics 55 56 57 58 59 60

65 66

517 – 526 527 – 532 533 – 544 545 – 548 549 – 552 553

554 – 585 Basics of Power Semiconductor Devices Phase Controlled Rectifier Choppers Inverters AC Voltage Regulators and Cycloconverters Applications of Power Electronics

554 – 559 560 – 570 571 – 575 576 – 581 582 583 – 585

#11. Electrical Machines 61 62 63 64

484 – 487 488 – 498 499 – 503 504 – 509 510 – 516

586 - 621

Transformer Induction Motor D.C. Machine Synchronous Machine Principles of Electro Mechanical Energy Conversion Special Machines

th

th

586 – 596 597 – 605 606 – 611 612 – 618 619 – 620 621

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page II

GATE QUESTION BANK

Mathematics

Linear Algebra ME – 2005 1. Which one of the following is an Eigenvector of the matrix[

(A) [

]

(B) [ ]

2.

5.

]?

(C) [

]

(D) [

]

A is a 3 4 real matrix and Ax=B is an inconsistent system of equations. The highest possible rank of A is (A) 1 (C) 3 (B) 2 (D) 4

ME – 2006 3. Multiplication of matrices E and F is G. Matrices E and G are os sin E [ sin ] and os G

4.

[

sin os

sin (B) [ os

os sin

]

os (C) [ sin

sin os

]

sin (D) [ os

os sin

7.

Eigenvectors of 0

1 is

(A) 0 (B) 1

(C) 2 (D) Infinite

If a square matrix A is real and symmetric, then the Eigenvalues (A) are always real (B) are always real and positive (C) are always real and non-negative (D) occur in complex conjugate pairs

]

ME – 2008 8.

The Eigenvectors of the matrix 0

1 are

written in the form 0 1 and 0 1. What is a + b? (A) 0 (B) 1/2

]

Eigen values of a matrix 0

ME – 2007 6. The number of linearly independent

]. What is the matrix F?

os (A) [ sin

S

Match the items in columns I and II. Column I Column II P. Singular 1. Determinant is not matrix defined Q. Non-square 2. Determinant is matrix always one R. Real 3. Determinant is symmetric zero matrix S. Orthogonal 4. Eigen values are matrix always real 5. Eigen values are not defined (A) P - 3 Q - 1 R - 4 S - 2 (B) P - 2 Q - 3 R - 4 S - 1 (C) P - 3 Q - 2 R - 5 S - 4 (D) P - 3 Q - 4 R - 2 S - 1

9.

(C) 1 (D) 2

The matrix [

] has one Eigenvalue p equal to 3. The sum of the other two Eigenvalues is (A) p (C) p – 2 (B) p – 1 (D) p – 3

1are 5 and 1. What are the

Eigenvalues of the matrix = SS? (A) 1 and 25 (C) 5 and 1 (B) 6 and 4 (D) 2 and 10 th

th

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 1

GATE QUESTION BANK

10.

For what value of a, if any, will the following system of equations in x, y and z have a solution x y x y z x y z (A) Any real number (B) 0 (C) 1 (D) There is no such value

11.

ME – 2012 15.

For a matrix,M-

*

x

√

(B) (√ )

1 is

(A) 2 (B) 2 3

3

(C) 2 3 (D) 2

3

ME – 2011 13. Consider the following system equations: x x x x x x x The system has (A) A unique solution (B) No solution (C) Infinite number of solutions (D) Five solutions 14.

of

Eigen values of a real symmetric matrix are always (A) Positive (C) Negative (B) Real (D) Complex

(D) ( ) √

√

of the matrix is equal to the inverse of the ,M- . The value of x is matrix ,Mgiven by ) (A) ( (C) ⁄ ( ⁄ ) (B) (D) ⁄

0

1 , one of the

(C) (√ )

(A) (√ )

+, the transpose

ME – 2010 12. One of the Eigenvectors of the matrix

For the matrix A=0

normalized Eigenvectors is given as

16.

ME – 2009

Mathematics

x + 2y + z =4 2x + y + 2z =5 x–y+z=1 The system of algebraic equations given above has (A) a unique algebraic equation of x = 1, y = 1 and z = 1 (B) only the two solutions of ( x = 1, y = 1, z = 1) and ( x = 2, y = 1, z = 0) (C) infinite number of solutions. (D) No feasible solution.

ME – 2013 17. The Eigenvalues of a symmetric matrix are all (A) Complex with non –zero positive imaginary part. (B) Complex with non – zero negative imaginary part. (C) Real (D) Pure imaginary. 18.

Choose correct set of functions, which are linearly dependent. (A) sin x sin x n os x (B) os x sin x n t n x (C) os x sin x n os x (D) os x sin x n os x

ME – 2014 19. Given that the determinant of the matrix [

] is

12 , the determinant of

the matrix [ (A) th

] is (B)

th

(C) th

(D)


GATE QUESTION BANK

20.

One of the Eigenvectors of the matrix 0

21.

22.

2.

Consider a non-homogeneous system of linear equations representing mathematically an over-determined system. Such a system will be (A) consistent having a unique solution (B) consistent having many solutions (C) inconsistent having a unique solution (D) inconsistent having no solution

3.

Consider the matrices , - . The order of , (

1 is

(A) {– }

(C) 2

(B) {– }

(D) 2 3

3

Consider a 3×3 real symmetric matrix S such that two of its Eigenvalues are with respective Eigenvectors x y [x ] [y ] If then x y + x y +x y x y equals (A) a (C) ab (B) b (D) 0 Which one of the following equations is a correct identity for arbitrary 3×3 real matrices P, Q and R? (A) ( ) ) (B) ( ( ) (C) et et et ) (D) (

CE – 2005 1. Consider the system of equations ( ) is s l r Let ( ) ( ) where ( ) e n Eigen -pair of an Eigenvalue and its corresponding Eigenvector for real matrix A. Let I be a (n × n) unit matrix. Which one of the following statement is NOT correct? (A) For a homogeneous n × n system of linear equations,(A ) X = 0 having a nontrivial solution the rank of (A ) is less than n. (B) For matrix , m being a positive integer, ( ) will be the Eigen pair for all i. (C) If = then | | = 1 for all i. (D) If = A then is real for all i.

Mathematics

,

-

,

-

and

- will be ) (C) (4 × 3) (D) (3 × 4

(A) (2 × 2) (B) (3 × 3

CE – 2006 4. Solution for the system defined by the set of equations 4y + 3z = 8; 2x – z = 2 and 3x + 2y = 5 is (A) x = 0; y =1; z = ⁄ (B) x = 0; y = ⁄ ; z = 2 (C) x = 1; y = ⁄ ; z = 2 (D) non – existent

5.

For the given matrix A = [

],

one of the Eigen values is 3. The other two Eigen values are (A) (C) (B) (D) CE – 2007 6. The minimum and the maximum Eigenvalue of the matrix [

]are 2

and 6, respectively. What is the other Eigenvalue? (A) (C) (B) (D) 7.

For what values of and the following simultaneous equations have an infinite of solutions? X + Y + Z = 5; X + 3Y + 3Z = 9; X+2Y+ Z (A) 2, 7 (C) 8, 3 (B) 3, 8 (D) 7, 2 th

th

th


GATE QUESTION BANK

8.

The inverse of the (A)

0

(B)

0

1 1

m trix 0

(A) (B)

1

is

(D)

0

1

( )

0

( )

0

( )

0

( )

0

11.

is

15. (C) (D)

i

i

i

i i

i

i

i i

i

i

i i

i i

1

1

1

1

i

i

i i

i

1

CE – 2012

1 are and 8 and 5

The inverse of the matrix 0

0

The Eigenvalue of the matrix [P] = 0

14.

(C)

CE – 2008 9. The product of matrices ( ) (A) (C) (B) (D) PQ 10.

1 is

Mathematics

n n

The following simultaneous equation x+y+z=3 x + 2y + 3z = 4 x + 4y + kz = 6 will NOT have a unique solution for k equal to (A) 0 (C) 6 (B) 5 (D) 7

CE – 2009 12. A square matrix B is skew-symmetric if (C) (A) (D) (B) CE – 2011 13. [A] is square matrix which is neither symmetric nor skew-symmetric and , is its transpose. The sum and difference of these matrices are defined as [S] = [A] + , - and [D] = [A] , - , respectively. Which of the following statements is TRUE? (A) Both [S] and [D] are symmetric (B) Both [S] and [D] are skew-symmetric (C) [S] is skew-symmetric and [D] is symmetric (D) [S] is symmetric and [D] is skew symmetric

The Eigenvalues of matrix 0 (A) (B) (C) (D)

1 are

2.42 and 6.86 3.48 and 13.53 4.70 and 6.86 6.86 and 9.50

CE – 2013 16. There is no value of x that can simultaneously satisfy both the given equations. Therefore, find the ‘le st squares error’ solution to the two equations, i.e., find the value of x that minimizes the sum of squares of the errors in the two equations. 2x = 3 and 4x = 1 17.

What is the minimum number of multiplications involved in computing the matrix product PQR? Matrix P has 4 rows and 2 columns, matrix Q has 2 rows and 4 columns, and matrix R has 4 rows and 1 column. __________

CE – 2014 18.

Given the matrices J = [ K

19.

[

] n

], the product K JK is

The sum of Eigenvalues of the matrix, [M] is, where [M] = [

]

(A) 915 (B) 1355 th

th

(C) 1640 (D) 2180 th


GATE QUESTION BANK

4. 20.

The determinant of matrix [

Let A be a 4x4 matrix with Eigenvalues –5, –2, 1, 4. Which of the following is an I Eigenvalue of 0 1, where I is the 4x4 I identity matrix? (A) (C) (B) (D)

]

is ____________ 21.

The

rank

[

of

the

matrix

] is ________________

CS – 2005 1. Consider the following system of equations in three real variables x x n x x x x x x x x x x This system of equation has (A) no solution (B) a unique solution (C) more than one but a finite number of solutions (D) an infinite number of solutions 2.

What are the Eigenvalues of the following 2 2 matrix? 0 (A) (B)

1 n n

(C) (D)

n n

CS – 2006 3. F is an n x n real matrix. b is an n real vector. Suppose there are two nx1 vectors, u and v such that u  v , and Fu=b, Fv=b. Which one of the following statement is false? (A) Determinant of F is zero (B) There are infinite number of solutions to Fx=b (C) There is an x  0 such that Fx=0 (D) F must have two identical rows

Mathematics

CS – 2007 5. Consider the set of (column) vectors defined by X={xR3 x1+x2+x3=0, where XT =[x1, x2, x3]T }. Which of the following is TRUE? (A) {[1, 1, 0]T, [1, 0, 1]T} is a basis for the subspace X. (B) {[1, 1, 0]T, [1, 0, 1]T} is a linearly independent set, but it does not span X and therefore, is not a basis of X. (C) X is not the subspace for R3 (D) None of the above CS – 2008 6. The following system of x x x x x x x x x Has unique solution. The only possible value (s) for is/ are (A) 0 (B) either 0 or 1 (C) one of 0,1, 1 (D) any real number except 5 7.

How many of the following matrices have an Eigenvalue 1? 0

1 0

1 n 0

1 0

(A) One (B) two

1

(C) three (D) four

CS – 2010 8. Consider the following matrix A=[

] x y If the Eigen values of A are 4 and 8, then (A) x = 4, y = 10 (C) x = 3, y = 9 (B) x = 5, y = 8 (D) x = 4, y = 10 th

th

th


GATE QUESTION BANK

CS – 2011 9. Consider the matrix as given below [

13.

The value of the dot product of the Eigenvectors corresponding to any pair of different Eigenvalues of a 4-by-4 symmetric positive definite matrix is __________.

14.

If the matrix A is such that

]

Which one of the following options provides the CORRECT values of the Eigenvalues of the matrix? (A) 1, 4, 3 (C) 7, 3, 2 (B) 3, 7, 3 (D) 1, 2, 3

[

CS – 2013 11. Which one of x x equal [ y y z z x(x y(y (A) | z(z x (B) | y z x y (C) | y z z x y (D) | y z z

15.

-

The product of the non – zero Eigenvalues of the matrix

is __________. [ 16.

the following does NOT ] ) x ) y | ) z x | y z x y y z | z x y y z | z

],

Then the determinant of A is equal to __________.

CS – 2012 10. Let A be the 2

2 matrix with elements and . Then the Eigenvalues of the matrix are (A) 1024 and (B) 1024√ and √ (C) √ n √ (D) √ n √

Mathematics

]

Which one of the following statements is TRUE about every n n matrix with only real eigenvalues? (A) If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative. (B) If the trace of the matrix is positive, all its eigenvalues are positive. (C) If the determinant of the matrix is positive, all its eigenvalues are positive. (D) If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.

ECE – 2005 1. Given an orthogonal matrix

CS – 2014 12. Consider the following system of equations: x y x z x y z x y z The number of solutions for this system is __________.

A= [

]. ,

-

is

⁄ (A) [

⁄

]

⁄ ⁄

th

th

th


GATE QUESTION BANK

Mathematics

⁄ ⁄

(B) [

6.

]

⁄

The rank of the matrix [

⁄ (C) [

(A) 0 (B) 1

] ⁄ ⁄

(D) [

]

⁄ ⁄

2.

Let,

A=0

1 and

Then (a + b)= (A) ⁄ (B) ⁄ 3.

= 0

1.

⁄ ⁄

(C) (D)

Given the matrix 0

⁄

Eigenvector is (C) 0

1

(B) 0 1

(D) 0

1

ECE – 2006 4.

For the matrix 0 corresponding 0

the

ECE – 2007 7. It is given that X1 , X2 …… M are M nonzero, orthogonal vectors. The dimension of the vector space spanned by the 2M vector X1 , X2 … XM , X1 , X2 … XM is (A) 2M (B) M+1 (C) M (D) dependent on the choice of X1 , X2 … XM.

9.

All the four entries of the 2 x 2 matrix p p P = 0p p 1 are non-zero, and one of its Eigenvalues is zero. Which of the following statements is true? (A) p p p p (B) p p p p (C) p p p p (D) p p p p

Eigenvector

1 is

(A) 2 (B) 4 5.

1 , the Eigenvalue to

(C) 6 (D) 8

The Eigenvalues and the corresponding Eigenvectors of a 2 2 matrix are given by Eigenvalue Eigenvector =8

v =0 1

=4

(C) 2 (D) 3

ECE – 2008 8. The system of linear equations 4x + 2y = 7, 2x + y = 6 has (A) a unique solution (B) no solution (C) an infinite number of solutions (D) exactly two distinct solutions

1 the

(A) 0 1

]

v =0

1

ECE – 2009 10. The Eigen values of the following matrix are [

The matrix is (A) 0

1

(C) 0

1

(B) 0

1

(D) 0

1

]

(A) 3, 3 + 5j, 6 j (B) 6 + 5j, 3 + j, 3 j (C) 3 + j, 3 j, 5 + j (D) 3, 1 + 3j, 1 3j

th

th

th


GATE QUESTION BANK

ECE – 2010 11. The Eigenvalues of a skew-symmetric matrix are (A) Always zero (B) Always pure imaginary (C) Either zero or pure imaginary (D) Always real ECE – 2011 12. The system of equations x y z x y z x y z has NO solution for values of given by (A) (C) (B) (D)

Mathematics

ECE – 2014 16. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (A) (M ) M (M) (B) ( M ) (C) (M N) M N (D) MN NM 17.

A real (4 × 4) matrix A satisfies the equation I where 𝐼 is the (4 × 4) identity matrix. The positive Eigenvalue of A is _____.

18.

Consider the matrix

n

J ECE\EE\IN – 2012 13.

Given that A = 0

1 and I = 0

the value of A3 is (A) 15 A + 12 I (B) 19A + 30

(C) 17 A + 15 I (D) 17A +21

ECE – 2013 14. The minimum Eigenvalue of the following matrix is [

19.

The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ________.

20.

The system of linear equations

]

(A) 0 (B) 1 15.

[ ] Which is obtained by reversing the order of the columns of the identity matrix I . Let I J where is a nonnegative real number. The value of for which det(P) = 0 is _____.

1,

(C) 2 (D) 3

(

Let A be a m n matrix and B be a n m matrix. It is given that ) determinant Determinant(I (I ) where I is the k k identity matrix. Using the above property, the determinant of the matrix given below is

(A) 2 (B) 5

(

)h s

(A) a unique solution (B) infinitely many solutions (C) no solution (D) exactly two solutions 21.

[

)4 5

] (C) 8 (D) 16

th

Which one of the following statements is NOT true for a square matrix A? (A) If A is upper triangular, the Eigenvalues of A are the diagonal elements of it (B) If A is real symmetric, the Eigenvalues of A are always real and positive th

th


GATE QUESTION BANK

(C) If A is real, the Eigenvalues of A and are always the same (D) If all the principal minors of A are positive, all the Eigenvalues of A are also positive 22.

The maximum value of the determinant among all 2×2 real symmetric matrices with trace 14 is ___.

EE – 2005 1.

5.

If R = [

] , then top row of

(A) , (B) ,

2.

-

(C) , (D) ,

(B) [

] [

] [

]

(C) [

] [

] [

]

(D) [

] [

] [

]

-

(A) [ ] (B) [

]

(C) [

(B) [

]

(D) [ ]

]

In the matrix equation Px = q, which of the following is necessary condition for the existence of at least one solution for the unknown vector x (A) Augmented matrix [P/Q] must have the same rank as matrix P (B) Vector q must have only non-zero elements (C) Matrix P must be singular (D) Matrix P must be square

] ,R=[

(C) [ ] ]

(D) [

]

EE – 2007 6. X = [x , x . . . . x - is an n-tuple non-zero vector. The n n matrix V = X (A) Has rank zero (C) Is orthogonal (B) Has rank 1 (D) Has rank n 7.

The linear operation L(x) is defined by the cross product L(x) = b x, where b =[0 1 0- and x =[x x x - are three dimensional vectors. The matrix M of this operation satisfies x L(x) = M [ x ] x Then the Eigenvalues of M are (A) 0, +1, 1 (C) i, i, 1 (B) 1, 1, 1 (D) i, i, 0

8.

Let x and y be two vectors in a 3 dimensional space and denote their dot product. Then the determinant xx xy det 0 y x yy 1 (A) is zero when x and y are linearly independent (B) is positive when x and y are linearly independent (C) is non-zero for all non-zero x and y (D) is zero only when either x or y is zero

EE – 2006 Statement for Linked Answer Questions 4 and 5.

4.

]

is

] , one of

(A) [

] ,Q=[

] [

-

For the matrix p = [

P=[

(A) [

The following vector is linearly dependent upon the solution to the previous problem

the Eigenvalues is equal to 2 . Which of the following is an Eigenvector?

3.

Mathematics

] are

three vectors An orthogonal set of vectors having a span that contains P,Q, R is th

th

th


GATE QUESTION BANK

Statement for Linked Questions 9 and 10. Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix. A=0

A satisfies the relation (A) A + 3 + 2 =0 2 (B) A + 2A + 2 = 0 (C) (A+ ) (A 2) = 0 (D) exp (A) = 0

10.

equals (A) 511 A + 510  (B) 309 A + 104  (C) 154 A + 155  (D) exp (9A)

EE – 2008 11. If the rank of a ( ) matrix Q is 4, then which one of the following statements is correct? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent columns (C) Q will be invertible (D) Q will be invertible 12.

13.

(A) A A+ A = A (B) (AA+ ) = A A+ 14.

The characteristic equation of a ( ) matrix P is defined as () = | P| =    =0 If I denotes identity matrix, then the inverse of matrix P will be (A) ( I) (B) ( I) (C) ( I) (D) ( I)

(C) A+ A =  (D) A A+ A = A+

Let P be a real orthogonal matrix. x⃗ is a real vector [x x - with length ⃗x (x x ) . Then, which one of the following statements is correct? (A) x⃗ x⃗ where at least one vector satisfies x⃗ x⃗ (B) x⃗ x⃗ for all vectors x⃗ (C) x⃗ x⃗ where at least one vector satisfies x⃗ x⃗ (D) No relationship can be established between x⃗ and x⃗

1

9.

Mathematics

EE – 2009 15. The trace and determinant of a matrix are known to be –2 and –35 respe tively It’s Eigenv lues re (A) –30 and –5 (C) –7 and 5 (B) –37 and –1 (D) 17.5 and –2 EE – 2010 16. For the set of equations x x x x =2 x x x x =6 The following statement is true (A) Only the trivial solution x x x x = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist

17.

An Eigenvector of

[

(A) , (B) ,

(C) , (D) ,

-

] is -

EE – 2011 18.

The matrix[A] = 0

1 is decomposed

into a product of a lower triangular matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are

A is m n full rank matrix with m > n and  is an identity matrix. Let matrix A+ = ( ) , then, which one of the following statements is FALSE?

(A) 0 th

th

1 and 0

1 th


GATE QUESTION BANK

(B) 0 (C) 0 (D) 0

1 and 0 1 and 0 1 and 0

23.

1 1 1

EE – 2013 19.

The equation 0

x 1 0x 1

0 1 has

0 1.

Eigenvector of the matrix A = 0

(C) Non – zero unique solution (D) Multiple solution 20.

(A) [ 1 1]T (B) [3 1]T

A matrix has Eigenvalues – 1 and – 2. The corresponding Eigenvectors are 0 0

1 respectively. The matrix is

(A) 0

1

(C) 0

1

(B) 0

1

(D) 0

1

Which one of the following statements is true for all real symmetric matrices? (A) All the eigenvalues are real. (B) All the eigenvalues are positive. (C) All the eigenvalues are distinct. (D) Sum of all the eigenvalues is zero.

1?

(C) [1 1]T (D) [ 2 1]T

Let A be a 3 3 matrix with rank 2. Then AX = 0 has (A) only the trivial solution X = 0 (B) one independent solution (C) two independent solutions (D) three independent solutions

2.

1 and

EE – 2014 21. Given a system of equations: x y z x y z Which of the following is true regarding its solutions? (A) The system has a unique solution for any given and (B) The system will have infinitely many solutions for any given and (C) Whether or not a solution exists depends on the given and (D) The system would have no solution for any values of and 22.

Two matrices A and B are given below: p q pr qs p q [ ] 0 1 r s pr qs r s If the rank of matrix A is N, then the rank of matrix B is (A) N (C) N (B) N (D) N

IN – 2005 1. Identify which one of the following is an

(A) No solution x (B) Only one solution 0x 1

Mathematics

IN – 2006 Statement for Linked Answer Questions 3 and 4 A system of linear simultaneous equations is given as Ax=B where [

] n

[ ]

3.

The rank of matrix A is (A) 1 (C) 3 (B) 2 (D) 4

4.

Which of the following statements is true? (A) x is a null vector (B) x is unique (C) x does not exist (D) x has infinitely many values

5.

For a given that 0

1

matrix A, it is observed 0

1 n

0

1

0

1

Then matrix A is

th

th

th


GATE QUESTION BANK

 2 1  1 0   1 1     1 1  0 2 1 2

10.

(A) A  

1

1  1 0  2

1

1  1

1

(B) A      1 2 0 2 1 1 (C) A    1 2  0

02 1 2 1 1

0 2

(D) A    1 3 IN – 2007 6. Let A = [ ] i j n with n = i. j. Then the rank of A is (A) (C) n (B) (D) n 7.

n

Let A be an n×n real matrix such that = I and y be an n- dimensional vector. Then the linear system of equations Ax=Y has (A) no solution (B) a unique solution (C) more than one but finitely many independent solutions (D) Infinitely many independent solutions

The matrix P =[

12.

9.

The Eigenvalues of a (2 2) matrix X are 2 and 3. The Eigenvalues of matrix ( I) ( I) are (A) (C) (B) (D)

(

)(

)

(D) n IN – 2011 13.

The matrix M = [

] has

Eigenvalues . An Eigenvector corresponding to the Eigenvalue 5 is , - . One of the Eigenvectors of the matrix M is (A) , (C) , √ (B) , (D) ,

] rotates a vector

(C) (D)

A real n × n matrix A = [ ] is defined as i i j follows: { otherwise The summation of all n Eigenvalues of A is (A) n(n ) (B) n(n ) (C)

about the axis[ ] by an angle of (A) (B)

Let P 0 be a 3 3 real matrix. There exist linearly independent vectors x and y such that Px = 0 and Py = 0. The dimension of the range space of P is (A) 0 (C) 2 (B) 1 (D) 3

IN – 2010 11. X and Y are non-zero square matrices of size n n. If then (A) |X| = 0 and |Y| 0 (B) |X| 0 and |Y| = 0 (C) |X| = 0 and |Y| = 0 (D) |X| 0 and |Y| 0

IN – 2009 8.

Mathematics

IN – 2013 14. The dimension of the null space of the

15.

matrix [

] is

(A) 0 (B) 1

(C) 2 (D) 3

One of Eigenvectors corresponding to the two Eigenvalues of the matrix 0 (A) [

j

] 0

(B) 0 1 0 th

th

j

1 is

(C) [ ] 0 1 j j (D) [ ] 0 1 j

1 1 th


GATE QUESTION BANK

Mathematics

IN – 2014 16. For the matrix A satisfying the equation given below, the eigenvalues are , -[

]

[

(A) ( 𝑗,𝑗) (B) (1,1,0)

] (C) ( ) (D) (1,0,0)

Answer Keys and Explanations ME 1.

and G = [

[Ans. A] [

Now E × F = G

]

h r teristi equ tions is | I| ( )( )( ) ∴ Real eigenvalues are 5, 5 other two are complex Eigenvector corresponding to is ( I) (or) →( ) Verify the options which satisfies relation (1) Option (A) satisfies. [Ans. B] Given

n

in onsistent

4.

5.

[Ans. A]

6.

[Ans. B] 1 Eigenv lues re 2, 2 I)

(

I)

No. of L.I Eigenvectors ( (no of v ri les)

( ⁄ )

7.

matrix be A = 0 sin os

.

/ I)

[Ans. A] ( I) . olving for , Let the symmetric and real

[Ans. C] os Given , E = [ sin

]

matrix, if Eigenvalues are … … … … … then for matrix, the Eigenvalues will be , , ……… For S matrix, if Eigenvalues are 1 and 5 then for matrix, the Eigenvalues are 1 and 25.

No (

3.

sin os

[Ans. A] For S

0

( ) n ( ⁄ ) ( ( ) minimum of m n) For inconsistence ( ) ( ⁄ ) ∴ he highest possi le r nk of is

os [ sin

,E-

∴

2.

]

]

th

1

Now |

|

Which gives ( ⟹

)

th

th


GATE QUESTION BANK

⟹ Hence real Eigen value. 8.

Mathematics

x [

][

]

0

1

x

[Ans. B] Let

0

eigenv lues re

1

12.

Eigen vector corresponding to is ( I) x . / .y/ . / By simplifying K . / . / y t king K

⁄

Equating the elements x

n

[Ans. A] 0

1 → Eigenv lues re

Eigenve tor is x 13.

Eigen vector corresponding to =2 is ( I) x . / .y/ . / K By simplifying ( ) 4 5 by ⁄ K

[Ans. C] [

]

[

→

[

]

→

taking K

x verify the options

( )

[

]

]

infinite m ny solutions

⁄ ⁄ 9.

10.

[Ans. C] Sum of the diagonal elements = Sum of the Eigenvalues ⟹ 1 + 0 + p = 3+S ⟹ S= p 2

[Ans. B] Eigenvalues of a real symmetric matrix are always real

15.

[Ans. B] 0

If

1 eigenv lues v lue

Eigen vector will be .

/

Norm lize ve tor

[Ans. B] ( ⁄ )

11.

14.

[

]

√( )

(

)

[

]

[√( )

(

) ]

→ →

[

→

[

*

]

]

16.

system will h ve solution

[Ans. A] iven M

M

→ MM

I

th

⁄ √ + ⁄ √

[Ans. C] The given system is x y z x y z x y z Use Gauss elimination method as follows Augmented matrix is

th

th


GATE QUESTION BANK

, | -

[

→

→

| ] [

So, |

|

[

|

|

[Ans. C] Suppose the Eigenvalue of matrix A is ( i )(s y) and the Eigenvector is ‘x’ where s the onjug te p ir of Eigenvalue and Eigenvector is ̅ n x̅. So Ax = x … ① and x̅ ̅x̅……② king tr nspose of equ tion ② x̅ x̅ ̅ … ③ [( ) n ̅ is s l r ] ̅ x̅ x x̅ x x̅ x x̅ ̅x … , x̅ x x̅ ̅ x ̅ (x̅ x) ( ̅ re s l r ) (x̅ x) ̅

20.

[Ans. C] We know that os x os x sin x ( ) os x sin x ( ) os x Hence 1, 1 and 1 are coefficients. They are linearly dependent.

1 eigen v lues

Eigenve tor is

verify for oth n

21.

[Ans. D] We know that the Eigenvectors corresponding to distinct Eigenvalues of real symmetric matrix are orthogonal. x y x y [ ][ ] x y x y x y y x

22.

[Ans. D] ( ) In case of matrix PQ

CE 1.

QP (generally)

[Ans. C] If = i.e. A is orthogonal, we can only s y th t if is n Eigenv lue of then

also will be an Eigenvalue of A,

which does not necessarily imply that | | = 1 for all i. 2.

[Ans. A] In an over determined system having more equations than variables, it is necessary to have consistent unique solution, by definition

3.

[Ans. A] With the given order we can say that order of matrices are as follows:  3×4 Y  4×3  3×3

[Ans. A] |

[Ans. D] 0

nnot e zero )

Hence Eigenvalue of a symmetric matrix are real

19.

|

(Taking 2 common from each row) ( )

]

( x x̅ re Eigenve tors they i i i 0

18.

|

]

nk ( ) nk ( | ) So, Rank (A) = Rank (A|B) = 2 < n (no. of variables) So, we have infinite number of solutions 17.

Mathematics

|

th

th

th


GATE QUESTION BANK

( )  3×3 P 2×3  3×2 P( )  (2×3) (3×3) (3×2) 2×2 ( ( ) )  2×2

Using Gauss elimination we reduce this to an upper triangular matrix to find its rank | ]→

[

→ 4.

[Ans. D] The augmented matrix for given system is [

| ]→

[

| ]→

| ]

[

8.

| ]

[

[

|

[

|

]

( ⁄ ) ( ) ( ) ( ⁄ ) ∴ olution is non – existent for above system. 5.

6.

7.

[Ans. B] ∑ = Trace (A) + + = Trace (A) = 2 + ( 1) + 0 = 1 Now = 3 ∴3+ + =1  Only choice (B) satisfies this condition. [Ans. B] ∑ = Trace (A) + + =1+5+1=7 Now = 2, = 6 ∴ 2+6+ =7 =3

0

1

∴0

]

1 is (

1

)

(

) 0

9.

10.

0

1

0

1 1

[Ans. B] ( ) P=( ( )( ) =( ) (I) =

)P

[Ans. B] A=0

1

Characteristic equation of A is |

|=0

 (4 )( 5 ) 2 × 5 =0  + 30 = 0 6, 5 11.

[Ans. A] The augmented matrix for given system is [

]

[Ans. A] Inverse of 0

→

|

Now for infinite solution last row must be completely zero ie –2=0 n –7=0  n

Then by Gauss elimination procedure [

Mathematics

| ]

th

[Ans. D] The augmented matrix for given system is x [ | ] 6y7 [ ] z k Using Gauss elimination we reduce this to an upper triangular matrix to find its rank

th

th


GATE QUESTION BANK

| ]→

[

17.

[Ans. 16] , , M trix , The product of matrix PQR is , - , - , The minimum number of multiplications involves in computing the matrix product PQR is 16

18.

[Ans. 23]

k [

| ]

[

| ]

→

Now if k Rank (A) = rank (A|B) = 3 ∴ Unique solution If k = 7, rank (A) = rank (A|B) = 2 which is less than number of variables ∴ When K = 7, unique solution is not possible and only infinite solution is possible 12.

[Ans. A] A square matrix B is defined as skewsymmetric if and only if = B

13.

[Ans. D] By definition A + is always symmetric is symmetri is lw ys skew symmetri is skew symmetri

Mathematics

[

][

]

[

,

K JK

]

-[ ,

]

,

[

] -

-

19.

[Ans. A] Sum of Eigenvalues = Sum of trace/main diagonal elements = 215 + 150 + 550 = 915

20.

[Ans. 88] The determinant of matrix is [

]

→

14.

[Ans. B] 1 =(

0

i

∴ 0

15.

i

i

i

,( =

0

0

)

i)( i i

[

1

→

1 i -

i) i i

0

i i

i i

1

[

1

[

]

1 Interchanging Column 1& Column 2 and taking transpose

Sum of the Eigenvalues = 17 Product of the Eigenvalues = From options, 3.48 + 13.53 = 17 (3.48)(13.53) = 47 16.

]

→

[Ans. B] 0

]

[

[Ans. 0.5] 0.5

]

|

th

th

|

th


GATE QUESTION BANK * (

)

= ( 21.

(

(

)+

= 1, 6 ∴ The Eigenvalues of A are 1 and 6

[Ans. 2] ]

3.

[Ans. D] Given that Fu =b and Fv =b If F is non singular, then it has a unique inverse. Now, u = b and v= b Since is unique, u = v but it is given th t u v his is contradiction. So F must be singular. This means that (A) Determinant of F is zero is true. Also (B) There are infinite number of solution to Fx= b is true since |F| = 0 (C) here is n su h the is also true, since X has infinite number of solutions., including the X = 0 solution (D) F must have 2 identical rows is false, since a determinant may become zero, even if two identical columns are present. It is not necessary that 2 identical rows must be present for |F| to become zero.

4.

[Ans. C] It is given that Eigenvalues of A is 5, 2, 1, 4 I Let P = 0 1 I Eigenvalues of P : | I| I | | I ( ) I I I Eigenvalue of P is ( 5 +1 ), ( 2+ 1), (1+ 1), (4+1 ), ( 5 1 ), ( 2 1 ),(1 1), (4 1) = 4, 1, 2, 5, 6, 3,0,3

5.

[Ans. B] |x X= {x x x = ,x x x - then,

→ [ ( )

(

)

( ) ]

( )

( )

[

] ( )

no. of non zero rows = 2

[Ans. B] The augmented matrix for the given system is [

| ]

Using elementary transformation on above matrix we get, [

| ]

→

⁄ | ] ⁄ ⁄

[

→

[

|

]

Rank ([A B]) = 3 Rank ([A]) = 3 Since Rank ([A B]) = Rank ([A]) = number of variables, the system has unique solution. 2.

[Ans. B] 0

1

The characteristic equation of this matrix is given by | I| |

)

)

[

CS 1.

)(

Mathematics

+

| th

th

th


GATE QUESTION BANK

{ [1, 1, 0]T , [1,0, 1 ]T } is a linearly independent set because one cannot be obtained from another by scalar multiplication. However (1, 1, 0) and (1,0, 1) do not span X, since all such combinations (x1, x2, x3) such that x1+ x2+ x3 =0 cannot be expressed as linear combination of (1, 1,0) and (1,0, 1) 6.

7.

Only one matrix has an Eigenvalue of 1 which is 0

| ] →

→

[

[

1

Correct choice is (A) 8.

[Ans. D] |

| x y ( )( y) When ( y) x y x When ( y) x y x x y Solving (1) & (2) x y

[Ans. D] The augmented matrix for above system is [

Mathematics

| ] | ]

x

( )

( )

Now as long as – 5 0, rank (A) =rank (A|B) =3 ∴ can be any real value except 5. Closest correct answer is (D).

9.

[Ans. A] The Eigenvalues of a upper triangular matrix are given by its diagonal entries. ∴ Eigenvalues are 1, 4, 3 only

[Ans. A]

10.

[Ans. D]

Eigenvalues of 0 |

1

0

| =0

Eigenvalues of 0 |

Eigenvalues of the matrix (A) are the roots of the characteristic polynomial given below.

=0,1 1

|

| =0 =0

1

|

(√ )

)( ) =0 = –1, 1

n

√

n ( √ ) n

1

n

| =0

( (

) )

√ Eigenvalues of A are √ respectively So Eigenvalues of

) =0 ) = i or 1 = 1 –i or 1 + i

Eigenvalues of 0

)( )(

(

|= 0

( (

|

(

= 0, 0

Eigenvalues of 0 |

1

√

) =0

th

th

n

√

th


GATE QUESTION BANK

11.

12.

[Ans. A] → p q nd Since 2 & 3rd columns have been swapped which introduces a –ve sign Hence (A) is not equal to the problem

[

]

→

[

] →

→

[

] →

16.

[

]

( ) ( ) no of v ri ∴ nique solution exists

14.

[ ] x x Let X = x e eigen ve tor x [x ] By the definition of eigenvector, AX = x x x x x x x x [ ] [x ] [x ] x x x x x x x x x x x x x x x x x x x x x x n x x x x x x (I) If s yx x x x x x x x x x (2) If Eigenv lue ∴ Three distinct eigenvalues are 0, 2, 3 Product of non zero eigenvalues = 2 × 3 = 6

]

→ →

les

[Ans. 0] The Eigenvectors corresponding to distinct Eigenvalues of real symmetric matrix are orthogonal

ECE 1.

2.

[Ans. A] If the trace or determinant of matrix is positive then it is not necessary that all eigenvalues are positive. So, option (B), (C), (D) are not correct

[Ans. C] Since, ,

] (

=I

16

0

[

-

[Ans. A] We know,

[Ans. 0]

| |

[Ans. 6] Let A =

[Ans. 1] x y x z x y z x y z ugmente m trix is [

13.

15.

Mathematics

0

7=0 1

0

1 1

1 b ,  a 60 10 1 1 21 7   a+b =   3 60 60 20

)

Or 2a 0.1b=0,  2a 

th

th

th


GATE QUESTION BANK

3.

[Ans. C]

8.

0

[Ans. B] Approach 1: Given 4x + 2y =7 and 2x + y =6

1

 (A I)=0  ( 4 ) (3 ) 2 4=0  2 + 20=0  = 5, 4

4 2 x  7 2 1y   6      0 0 x  5 2 1y   6      

 x1   x2 

Putting = 5, 0

1   =0

 x + 2x = 0 x = 2x

On comparing LHS and RHS 0= 5, which is irrelevant and so no solution. Approach 2: 4x + 2y =7

x x  1= 2 2 1 Hence, 0 4.

1 is Eigenvector.

[Ans. C]

Then Eigenvector is x Verify the options (C) 5.

or 2x  y=

1 We know th t it is Eigenvalue

0

We know

0

1

|I A|=0

|

|

 2 –I2 +32 =0  = 4, 8 (Eigenvalues) For

= 4, ( I

)=0

1

)=0

1

9.

[Ans. C] Matrix will be singular if any of the Eigenvalues are zero. |  |= 0 For = 0, P = 0 p p  |p p | =0  p p p p

10.

[Ans. D] Approach1: Eigenvalues exists as complex conjugate or real Approach 2: Eigenvalues are given by

v =0 1 For

= 8, ( I

v =0 6.

1

[Ans. C] [

] [

|

]

[Ans. C] There are M non-zero, orthogonal vectors, so there is required M dimension to represent them ’

| =0

( 

( ) 7.

7 2

2x+y=6 Since both the linear equation represent parallel set of straight lines, therefore no solution exists. Approach 3: Rank (A)=1; rank (C)=2, As Rank (A) rank (C) therefore no solution exists.

x

[Ans. A] or m trix

Mathematics

11.

th

)(( ,

)=0

) j

j

[Ans. C] Eigenvalue of skew – symmetric matrix is either zero or pure imaginary. th

th


GATE QUESTION BANK

12.

13.

[Ans. B] Given equations are x y z x y z and x y z If and , then x y z have Infinite solution If and , then x y z ( ) no solution x y z If n x y z will have solution x y z and will also give solution

et of , -

et of [

]

16.

[Ans. D] Matrix multiplication is not commutative in general.

17.

[Ans. *] Range 0.99 to 1.01 Let ‘ ’ e Eigenv lue of ‘ ’ hen ‘ e Eigenv lue of ‘ ’ A. =I= Using Cauchey Hamilton Theorem,

[Ans. B] 0

Mathematics

’ will

1

Characteristic Equations is 18. By Cayley Hamilton theorem I ∴ ( I) I 14.

I | | [

[Ans. A] [

]

→

(

[

[Ans. *] Range 199 to 201 From matrix properties we know that the determinant of the product is equal to the product of the determinants. That is if A and B are two matrix with determinant | | n | | respectively, then | | | | | | ∴| | | | | |

20.

[Ans. B]

) ]

]

19.

| |

| | Product of Eigenvalues = 0 ∴ Minimum Eigenv lue h s to e ‘ ’ 15.

[Ans. *] Range 0.99 to 1.01 I J I J

[Ans. B] ,

Let

-

[ ]

[

I

I

[

Then AB = [4]; BA Here m = 1, n = 4 ) And et(I

]

]

→

[

[

[

]

th

les

[Ans. B] onsi er

)

]

]

( ) ( | ) no of v r Infinitely many solutions 21.

et(I

→ →

th

0

1 th


GATE QUESTION BANK

whi h is re l symmetri m trix h r teristi equ tion is | I| ( ) ∴ (not positive) ( ) is not true (A), (C), (D) are true using properties of Eigenvalues 22.

EE 1.

2.

[Ans. B] ] j( ) | |

=[

]

∴ Top row of

=,

-

[Ans. D] Since matrix is triangular, the Eigenvalues are the diagonal elements themselves namely  = 3, 2 & 1. Corresponding to Eigenvalue  = 2, let us find the Eigenvector [A - ] x̂ = 0 x [ ][x ] [ ] x Putting in above equation we get, x [ ][x ] [ ] x Which gives the equations, 5x x x =0 . . . . . (i) x =0 . . . . . (ii) 3x = 0 . . . . . (iii) Since eqa (ii) and (iii) are same we have 5x x x =0 . . . . . (i) x =0 . . . . . (ii) Putting x = k, we get x = 0, x = k and 5x k =0

[Ans. *] Range 48.9 to 49.1 Real symmetric matrices are diagnosable Let the matrix be x 0 1 s tr e is x So determinant is product of diagonal entries So | | x x ∴ M ximum v lue of etermin nt x x ∴| |

R= [

Mathematics

, of tor( )| |

x = k | |=|

|

∴ Eigenvectorss are of the form x k x [ ] * k + x

= 1(2 + 3) – 0(4 + 2) – 1 (6 – 2) = 1 Since we need only the top row of , we need to find only first column of (R) which after transpose will become first row adj(A). cof. (1, 1) = + |

|=2+3=5

cof. (2, 1) =

|= 3

|

cof. (2, 1) = + |

i.e. x x x = k : k : 0 = :1:0 =2:5:0 x x ∴ [ ]=[ ] is an Eigenvector of matrix p. x

|= +1 3.

∴ cof. (A) = [

[Ans. A] Rank [P|Q] = Rank [P] is necessary for existence of at least one solution to x q.

]

Adj (A) =, of ( )=[

]

Dividing by |R| = 1 gives th

th

th


GATE QUESTION BANK

4.

  (

[Ans. A] We need to find orthogonal vectors, verify the options. Option (A) is orthogonal vectors (

)(

[Ans. B] The vector ( ) is linearly dependent upon the solution obtained in - and , Q. No. 4 namely , We can easily verify the linearly dependence as |

6.

7.

i

[Ans. B] xy xx | yx

xy xx x n xy yx xy x xy y y | |y x y | (x y) x y = Positive when x and y are linearly independent.

Option (B), (C), (D) are not orthogonal 5.

) i

8.

)

Mathematics

9.

[Ans. A] A=0

1

|A – | = 0  |

|

[Ans. B] hen n n m trix xx x x x x x x x x x x x x * + x x x x x x Take x common from 1st row, x common from 2nd row …… x common from nth row. It h s r nk ‘ ’

| =0  A will satisfy this equation according to Cayley Hamilton theorem i.e. I=0 Multiplying by on oth si es we get I=0 I =0 10.

[Ans. A] To calculate Start from derived above

I = 0 which has I

[Ans. D] ⃗ k L(x) = |

(

| x

x

= (x )

I)(

x (

⃗( k

)

(

x )

I) I

x = x

⃗ =[ x k

(

x L(x) = M [x ] x Comparing both , we get,

(

|

)

I

I) (

I) I

| (

I)

I) I

(

Hence Eigenvalue of M : | M



I)( I

]

|

I

] x

M=[

I) I

(

) th

th

th


GATE QUESTION BANK

11.

12.

13.

[Ans. A] If rank of (5 6 ) matrix is 4,then surely it must have exactly 4 linearly independent rows as well as 4 linearly independent columns.

= A is correct =A[( ) -A = A[( ) Put =P Then A [ ] = A. = A Choice (C) =  is also correct since =( ) = I 14.

os

x in )

|| x⃗ || = √x

(x in

x

x

→

[Ans. C] Trace = Sum of Principle diagonal elements.

16.

[Ans. D] On writing the equation in the form of AX =B

+

, *

+

nk ( ) nk( ) Number of variables = 4 Since, Rank (A) = Rank(C) < Number of variables Hence, system of equations are consistent and there is multiple non-trivial solution exists. 17.

[Ans. B] Characteristic equation | |

I|

|

 (1 ) ( )( )  Eigenve tors orrespon ing to ( I) x [ ] [x ] [ ] x 2x x  x x At x x x x  x x At x ,x

is

Eigenvectors = c[ ]{Here c is a constant}

os ) 18.

[Ans. D] , - ,L-, - ⟹ Options D is correct

19.

[Ans. D] x x … (i) } (i) n (ii) re s me x x … (ii) ∴x x So it has multiple solutions.

|| x⃗ || = || x̅|| for any vector x̅ 15.

* +

Argument matrix C =*

[Ans. B] Let orthogonal matrix be os in P=0 1 in os By Property of orthogonal matrix A I x os x in So, x⃗ = [ ] x in x os || x⃗ || = √(x

x x + *x + x

*

[Ans. D] If characteristic equation is    =0 Then by Cayley – Hamilton theorem, I=0 = Multiplying by on both sides, = I = ( I) [Ans. D] Choice (A) Since

Mathematics

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GATE QUESTION BANK

20.

[Ans. D] Eigen value

|A

Eigenvectors 0

1 n 0

Let matrix 0 x

1

x 10

1

0

1

0

10

1

0

1

I|= |

|

i.e., (1 ) (2 ) 2 Thus the Eigenvalue are 1, 2. If x, y, be the component of Eigenvectors corresponding to the Eigenv lues we have x [A- I- 0 1 0y1=0

1

0

Mathematics

For =1, we get the Eigenvector as 0 Hence, the answer will be ,

21.

22.

23.

IN 1.

1

0

[Ans. B] AX=0 and (A) = 2 n=3 No. of linearly independent solutions = n r = 3 =1

3.

[Ans. C] There are 3 non-zero rows and hence rank (A) = 3

4.

[Ans. C] Rank (A) = 3 (This is Co-efficient matrix) Rank (A:b) =4(This is Augmented matrix) s r nk( ) r nk ( ) olution oes not exist.

5.

[Ans. C] We know Hen e from the given problem, Eigenvalue & Eigenvector is known.

1

[Ans. B] Since there are 2 equations and 3 variables (unknowns), there will be infinitely many solutions. If if then x y z x y z x z y For any x and z, there will be a value of y. ∴ Infinitely many solutions [Ans. A] For all real symmetric matrices, the Eigenvalues are real (property), they may be either ve or ve and also may be same. The sum of Eigenvalues necessarily not be zero. [Ans. C] p q 0 1 r s ( pplying → p q →r s element ry tr nsform tions) p q pr qs [ ] pr qs r s ∴ hey h ve s me r nk N

1 X1    , X2  1

1 2 , 1  1, 2  2  

We also know that

, where

1 1

P  X1 X2     1 2

1 0  1 0     0 2   0 2

[Ans. B] Given:

-

2. Solving 0

1

& D=  0

1 Hence

Characteristic equation is,

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GATE QUESTION BANK

 1 1  1 0   2 1  A       1 2  0 2 1 1 6.

12.

[Ans. B] A= [

]=[

A=[

[Ans. B] Given I Hence rank (A) = n Hence AX= Y will have unique solution

8.

[Ans. C]

9.

[Ans. C] Approach 1:

13.

14.

Assume,

0

(

∴A

(

0

1

0

10

Now | I

0

[Ans. B] Dim of null space [A]= nullity of A.

0

[

]

1

| )(

]

Apply row operations 1

1

- is also vector

For given A = [

1

I) 0

[Ans. B] If AX = → From this result [1, 2, for M

|

| (

I

1

I)

]

n For diagonal matrix Eigenvalues are diagonal elements itself. n(n ) ∴ n

]

Hence, rank (A) =1 7.

[Ans. A] A=[ ] i if i j = 0 otherwise. For n n matrix

]

Using elementary transformation [

Mathematics

)=0

[Ans. D]

11.

[Ans. C] A null matrix can be obtained by multiplying either with one null matrix or two singular matrices.

[

→

[

] ]

∴ ( ) By rank – nullity theorem Rank [A]+ nullity [A]= no. of columns[A] Nullity [A]= 3 ∴ Nullity , -

Approach 2: Eigenvalues of ( I) is = 1, 1/2 Eigenvalues of (X+5I) is = 3, 2 Eigenvalues of ( I) (X+5I) is = , 10.

→

15.

[Ans. A] A=|

|

Characteristics equation | |

I|

| j j

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GATE QUESTION BANK

j

[

j

x ] 0x 1

Mathematics

0 1

x x

j j

[

j j

x ] 0x 1

x

0 1

j

x 16.

[Ans. C]

A[

]=[

→| | |

|

] |

|

→| | (

|

|

|

| two rows ounter lose thus | |

| |) =Product of eigenvalues Verify options Options (C) correct answer

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th

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GATE QUESTION BANK

Mathematics

Probability and Distribution ME - 2005 1. A single die is thrown twice. What is the probability that the sum is neither 8 nor 9? (A) ⁄ (C) ⁄ ⁄ (B) ⁄ (D)

ME - 2008 6. A coin is tossed 4 times. What is the probability of getting heads exactly 3 times? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

2.

ME - 2009 7. The standard deviation of a uniformly distributed random variable between 0 and 1 is (A) (C) ⁄√ √ (B) (D) √ √

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is (A) 0.0036 (C) 0.2234 (B) 0.1937 (D) 0.3874

ME - 2006 3. Consider a continuous random variable with probability density function f(t) = 1 + t for 1  t  0 = 1 t for 0  t  1 The standard deviation of the random variable is: (C) ⁄ (A) ⁄√ (D) ⁄ (B) ⁄√ 4.

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective? ⁄ (A) ⁄ (C) ⁄ ⁄ (B) (D)

ME - 2007 5. Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE? (A) E (XY) = E (X) E (Y) (B) Cov (X, Y) = 0 (C) Var (X + Y) = Var (X) + Var (Y) (D)

(X Y )

( (X)) ( (Y))

8.

If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (C) 1/2 (B) 3/8 (D) 7/8

ME - 2010 9. A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is (A) 2/315 (C) 1/1260 (B) 1/630 (D) 1/2520 ME - 2011 10. An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is________ ⁄ (A) ⁄ (C) ⁄ ⁄ (B) (D) ME - 2012 11. A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set has one red ball and two black balls is (A) 1/20 (C) 3/10 (B) 1/12 (D) 1/2 th

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GATE QUESTION BANK

ME - 2013 12. Let X be a normal random variable with mean 1 and variance 4. The probability (X ) is (A) 0.5 (B) Greater than zero and less than 0.5 (C) Greater than 0.5 and less than 1.0 (D) 1.0 13.

The probability that a student knows the correct answer to a multiple choice

the probability of obtaining red colour on top face of the dice at least twice is _______ 17.

A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is _______

18.

A machine produces 0, 1 or 2 defective pieces in a day with associated probability of 1/6, 2/3 and 1/6, respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are (A) 1 and 1/3 (C) 1 and 4/3 (B) 1/3 and 1 (D) 1/3 and 4/3

19.

A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs. 500 is _______

20.

The number of accidents occurring in a plant in a month follows Poisson distribution with mean as 5.2. The probability of occurrence of less than 2 accidents in the plant during a randomly selected month is (A) 0.029 (C) 0.039 (B) 0.034 (D) 0.044

question is . If the student dose not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is . Given that the student has answered the questions correctly, the conditional probability that the student knows the correct answer is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ ME - 2014 14. In the following table x is a discrete random variable and P(x) is the probability density. The standard deviation of x is x 1 2 3 P(x) 0.3 0.6 0.1 (A) 0.18 (C) 0.54 (B) 0.3 (D) 0.6 15.

16.

Box contains 25 parts of which 10 are defective. Two parts are being drawn simultaneously in a random manner from the box. The probability of both the parts being good is ( )

( )

( )

( )

Consider an unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice. If the dice is thrown thrice,

Mathematics

CE - 2005 1. Which one of the following statements is NOT true? (A) The measure of skewness is dependent upon the amount of dispersion

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GATE QUESTION BANK

(B) In a symmetric distribution the value of mean, mode and median are the same (C) In a positively skewed distribution mean > median > mode (D) In a negatively skewed distribution mode > mean > median CE - 2006 2. A class of first years B. Tech students is composed of four batches A, B, C and D each consisting of 30 students. It is found that the sessional marks of students in Engineering Drawing in batch C have a mean of 6.6 and standard deviation of 2.3. The mean and standard deviation of the marks for the entire class are 5.5 and 4.2 respectively. It is decided by the course instruction to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from 8.5 to (A) 6.0 (C) 8.0 (B) 7.0 (D) 9.0 3.

There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e. each has the same chance of being selected). What is the probability that only one of the defective calculators will be included in the inspection? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

CE - 2007 4. If the standard deviation of the spot speed of vehicles in a highway is 8.8 kmph and the mean speed of the vehicles is 33 kmph, the coefficient of variation in speed is (A) 0.1517 (C) 0.2666 (B) 0.1867 (D) 0.3646

Mathematics

CE - 2008 5. If probability density function of a random variable x is x for x nd f(x) { for ny other v lue of x Then, the percentage probability P.

x

/ is

(A) 0.247 (B) 2.47 6.

(C) 24.7 (D) 247

A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using the public transport, further choices available are bus and metro out of which the probability of commuting by a bus is 0.55. In such a situation, the probability, (rounded upto two decimals) of using a car, bus and metro, respectively would be (A) 0.45, 0.30 and 0.25 (B) 0.45, 0.25 and 0.30 (C) 0.45, 0.55 and 0.00 (D) 0.45, 0.35 and 0.20

CE - 2009 7. The standard normal probability function can be approximated as (x )

|x | ) exp( Where x = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is (A) 66.7% (C) 33.3% (B) 50.0% (D) 16.7% CE - 2010 8. Two coins are simultaneously tossed. The probability of two heads simultaneously appearing is (A) 1/8 (C) 1/4 (B) 1/6 (D) 1/2

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GATE QUESTION BANK

CE - 2011 9. There are two containers with one containing 4 red and 3 green balls and the other containing 3 blue and 4 green balls. One ball is drawn at random from each container. The probability that one of the balls is red and the other is blue will be (A) 1/7 (C) 12/49 (B) 9/49 (D) 3/7 CE - 2012 10. The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is (A) < 50 % (C) 75 % (B) 50 % (D) 100 % 11.

14.

A traffic office imposes on an average 5 number of penalties daily on traffic violators. Assume that the number of penalties on different days is independent and follows a poisson distribution. The probability that there will be less than 4 penalties in a day is ____.

15.

A fair (unbiased) coin was tossed four times in succession and resulted in the following outcomes: (i) Head (iii) Head (ii) Head (iv) Head The prob bility of obt ining ‘T il’ when the coin is tossed again is (A) 0 (C) ⁄ (B) ⁄ (D) ⁄

16.

An observer counts 240 veh/h at a specific highway location. Assume that the vehicle arrival at the location is Poisson distributed, the probability of having one vehicle arriving over a 30-second time interval is ____________

In an experiment, positive and negative values are equally likely to occur. The probability of obtaining at most one negative value in five trials is (A)

(C)

(B)

(D)

CE - 2013 12. Find the value of such that the function f(x) is a valid probability density function ____________________ (x )( f(x) x) for x otherwise CE - 2014 13. The probability density function of evaporation E on any day during a year in a watershed is given by f( )

{

mm d y

Mathematics

CS - 2005 1. Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are 1/3 and 2/3 respectively. Given that a ball selected in the above process is red, the probability that it comes from box P is (A) 4/19 (C) 2/9 (B) 5/19 (D) 19/30 2.

Let f(x) be the continuous probability density function of a random variable X. The probability that a  X  b , is (A) f(b a) (C) ∫ f(x)dx

otherwise The probability that E lies in between 2 and 4 mm/day in a day in the watershed is (in decimal) ______

(B) f(b)

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f( )

(D) ∫ x f(x)dx

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GATE QUESTION BANK

CS - 2006 3. For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is (A) ( n ⁄ ) (C) ( ⁄ n ) (D) ⁄ (B) ( n ⁄ ) CS - 2007 Linked Data for Q4 & Q5 are given below. Solve the problems and choose the correct answers. Suppose that robot is placed on the Cartesian plane. At each step it is easy to move either one unit up or one unit right, i.e if it is at (i,j) then it can move to either (i+1,j) or (i,j+1) 4. How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)? (C) 210 (A) 20 (D) None of these (B) 2 5.

Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)? (A) 29 (B) 219 (C) . / . (D) .

6.

/

/ . / .

/

Suppose we uniformly and randomly select a permutation from the 20! ermut tions of ………… Wh t is the probability that 2 appears at an earlier position than any other even number in the selected permutation? (A) ⁄ (C) ⁄ (B) ⁄ (D) none of these

Mathematics

CS - 2008 7. Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean of 1 and variance unknown If (X ) (Y≥ ) the standard deviation of Y is (A) 3 (C) √ (B) 2 (D) 1 8.

Aishwarya studies either computer science or mathematics every day. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday? (A) 0.24 (C) 0.4 (B) 0.36 (D) 0.6

CS - 2009 9. An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3? (A) 0.453 (C) 0.485 (B) 0.468 (D) 0.492 CS - 2010 10. Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty? th

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GATE QUESTION BANK

(A) (B) (C) (D) 11.

12.

pq+(1 – p)(1 – q) (1 – q)p (1 – p)q pq

What is the probability that a divisor of is a multiple of ? (A) 1/625 (C) 12/625 (B) 4/625 (D) 16/625 If the difference between the expectation of the square if a random variable ( ,x -) and the square if the exopectation of the random variable ( ,x-) is denoted by R, then (A) R = 0 (C) R≥ (B) R< 0 (D) R > 0

CS - 2011 13. A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card? (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ 14.

Consider a finite sequence of random values X = [x1, x2 … xn].Let be the me n nd σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi, a*xi+b, where a and b are positive constants. Let μy be the me n nd σy be the standard deviation of this sequence. Which one of the following statements is INCORRECT? (A) Index position of mode of X in X is the same as the index position of mode of Y in Y. (B) Index position of median of X in X is the same as the index position of median of Y in Y. (C) μy μx + b (D) σy σx + b

15.

Mathematics

If two fair coins flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads? (A) 1/3 (C) 1/4 (B) 1/2 (D) 2/3

CS - 2012 16. Suppose a fair six – sided die is rolled once. If the value on the die is 1,2, or 3 the die is rolled a second time. What is the probability that the some total of value that turn up is at least 6? (A) 10/21 (C) 2/3 (B) 5/12 (D) 1/6 17.

Consider a random variable X that takes values +1 and 1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = and +1 are (A) 0 and 0.5 (C) 0.5 and 1 (B) 0 and 1 (D) 0.25 and 0.75

CS - 2013 18. Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? ⁄ e (A) ⁄ e (C) ⁄ e (B) ⁄ e (D) CS - 2014 19. Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ . 20.

th

Four fair six – sided dice are rolled. The probability that the sum of the results being 22 is x/1296. The value of x is ____________

th

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GATE QUESTION BANK

21.

The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p = _____________.

22.

Each of the nine words in the sentence “The quick brown fox jumps over the l zy dog” is written on sep r te piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)

23.

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is __________.

24.

Let S be a sample space and two mutually exclusive events A and B be such that ∪ S If ( ) denotes the prob bility of the event, the maximum value of P(A) P(B) is _______

ECE - 2006 3. A probability density function is of the ). form (x) e || x ( The value of K is (A) 0.5 (C) 0.5a (B) 1 (D) A 4.

Three Companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below Company % of Probability computers of being supplied defective X 60% 0.01 Y 30% 0.02 Z 10% 0.03 Given that a computer is defective, the probability that it was supplied by Y is (A) 0.1 (C) 0.3 (B) 0.2 (D) 0.4

ECE - 2007 5. If E denotes expectation, the variance of a random variable X is given by (A) E[X2] E2[X] (C) E[X2] (B) E[X2] + E2[X] (D) E2[X] 6.

An examination consists of two papers, Paper1 and Paper2. The probability of failing in Paper1 is 0.3 and that in Paper2 is 0.2.Given that a student has failed in Paper2, the probability of failing in paper1 is 0.6. The probability of a student failing in both the papers is (A) 0.5 (C) 0.12 (B) 0.18 (D) 0.06

ECE - 2005 1. A fair dice is rolled twice. The probability that an odd number will follow an even number is

2.

( )

( )

( )

( )

Mathematics

The value of the integral

I

 x2  1  exp   dx is 2 0  8

(A) 1 (B)

(C) 2 (D) th

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GATE QUESTION BANK

ECE - 2008 7. The probability density function (PDF) of a random variable X is as shown below.

(x) exp( |x|) exp( |x|) is the probability density function for the real random variable X, over the entire x axis. M and N are both positive real numbers. The equation relating M and N is

8.

PDF PDF

1

Mathematics

(A) 1

0

(B) 2M

x 11

The -1 corresponding cumulative 0 distribution function (CDF) has the form

(A)

1

(C) M + N = 1 (D) M + N = 3 ECE - 2009 9. Consider two independent random variables X and Y with identical distributions. The variables X and Y take value 0, 1 and 2 with probabilities

CDF

1

N=1

x

and respectively. What is the 1

1

0

(B)

x

conditional probability (x y ) |x y| (A) 0 (C) ⁄ ⁄ (B) (D) 1

CD F C

1

D F

1

10. 0

1 -1

(C)

x

1

2

(B)

11. 1

0

x

1

1 1 1

0 0 1

1  2

(C)   2

0

(D)

10

1  2

(A)  

CDF 1

0

1

A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?

CDF

1 1

x

th

10

10

1 C2    2

(D)

10

1 C2    2

A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean of X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true? k P(X=k) 1 0.1 2 0.2 3 0.4 4 0.2 5 0.1

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GATE QUESTION BANK

(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong ECE - 2010 12. A fair coin is tossed independently four times. The prob bility of the event “the number of times heads show up is more th n the number of times t ils show up” is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ ECE - 2011 13. A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (A) 2/36 (C) 5/12 (B) 2/6 (D) 1/2 ECE\EE\IN - 2012 14. A fair coin is tossed till a head appears for the first time probability that the number of required tosses is odd , is (A) 1/3 (C) 2/3 (B) 1/2 (D) 3/4 ECE - 2013 15. Let U and V be two independent zero mean Gaussian random variables of variances ⁄ and ⁄ respectively. The probability ( V ≥ U) is (A) 4/9 (C) 2/3 (B) 1/2 (D) 5/9 16.

Consider two identically distributed zeromean random variables U and V . Let the cumulative distribution functions of U and 2V be F(x) and G(x) respectively. Then, for all values of x (x)) (A) ( (x) (B) ( (x)

(C) ( (x) (D) ( (x)

Mathematics

(x)) x (x)) x ≥

ECE - 2014 17. In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is _____ 18.

Let X X nd X , be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X is the largest} is _____

19.

Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E[X], is __________.

20.

An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is (A) 0.067 (C) 0.082 (B) 0.073 (D) 0.091

21.

A fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is _______.

22.

Let X X and X be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability P{X X X } is ______.

23.

Let X be a zero mean unit variance Gaussian random variable. ,|x|- is equal to __________

24.

Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-office has a probability

of

losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is ____________.

(x)) ≥ th

th

th


GATE QUESTION BANK

EE - 2005 1. If P and Q are two random events, then the following is TRUE (A) Independence of P and Q implies that probability (P Q) = 0 (B) Probability (P ∪ Q)≥ Probability (P) +Probability (Q) (C) If P and Q are mutually exclusive, then they must be independent (D) Probability (P Q) Probability (P) 2.

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

EE - 2006 3. Two f ir dice re rolled nd the sum “ r ” of the numbers turned up is considered (A) Pr (r > 6) = (B) Pr (r/3 is an integer) = (C) Pr (r = 8|r/4 is an integer) = (D) Pr (r = 6|r/5 is an integer) = EE - 2007 4. A loaded dice has following probability distribution of occurrences Dice Value Probability ⁄ 1 2

⁄

3

⁄

4

⁄

5

⁄

⁄ 6 If three identical dice as the above are thrown, the probability of occurrence of values, 1, 5 and 6 on the three dice is (A) same as that of occurrence of 3, 4, 5 (B) same as that of occurrence of 1, 2, 5 (C) 1/128 (D) 5/8

Mathematics

EE - 2008 5. X is a uniformly distributed random variable that takes values between 0 and 1. The value of E{X } will be (A) 0 (C) 1/4 (B) 1/8 (D) 1/2 EE - 2009 6. Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of atleast two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5? (A) 20 (C) 15 (B) 7 (D) 16 EE - 2010 7. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is (A) 1/3 (C) 1/2 (B) 3/7 (D) 4/7 ECE\EE\IN - 2012 8. Two independent random variables X and Y are uniformly distributed in the interval , -. The probability that max , - is less than 1/2 is (A) 3/4 (C) 1/4 (B) 9/16 (D) 2/3 EE - 2013 9. A continuous random variable x has a probability density function + is f(x) e x . Then *x (A) 0.368 (C) 0.632 (B) 0.5 (D) 1.0

th

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GATE QUESTION BANK

EE - 2014 10. A fair coin is tossed n times. The probability that the difference between the number of heads and tails is (n – 3) is (C) (A) (B) (D) 11.

12.

13.

14.

IN - 2005 1. The probability that there are 53 Sundays in a randomly chosen leap year is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄ 2.

A mass of 10 kg is measured with an instrument and the readings are normally distributed with respect to the mean of 10 kg. Given that

Consider a dice with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is _______________ Let x be a random variable with probability density function for |x| f(x) { |x| for otherwise The probability P(0.5 < x < 5) is_________ Lifetime of an electric bulb is a random variable with density f(x) kx , where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is__________ The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are (A) 30 mm and 0.22 (B) 30 mm and 2.44 (C) 40 mm and 2.44 (D) 40 mm and 0.24

Mathematics

exp .

∫

√

/ d =0.6

and that 60per cent of the readings are found to be within 0.05 kg from the mean, the standard deviation of the data is (A) 0.02 (C) 0.06 (B) 0.04 (D) 0.08 3.

The measurements of a source voltage are 5.9V, 5.7V and 6.1V. The sample standard deviation of the readings is (A) 0.013 (C) 0.115 (B) 0.04 (D) 0.2

IN - 2006 4. You have gone to a cyber-cafe with a friend. You found that the cyber-café has only three terminals. All terminals are unoccupied. You and your friend have to make a random choice of selecting a terminal. What is the probability that both of you will NOT select the same terminal? (A) ⁄ (C) ⁄ (B) ⁄ (D) 1 5.

Probability density function p(x) of a random variable x is as shown below. The value of  is p(x) α

0

th

α

α b

α c

(A)

c

(C)

(B)

c

(D)

th

th

(

)

(

)


GATE QUESTION BANK

6.

Mathematics

Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even is (A) 0.5 (C) 0.167 (B) 0.25 (D) 0.125

measurements, it can be expected that the number of measurement more than 10.15 mm will be (A) 230 (C) 15 (B) 115 (D) 2

IN - 2007 7. Assume that the duration in minutes of a telephone conversation follows the

IN - 2011 12. The box 1 contains chips numbered 3, 6, 9, 12 and 15. The box 2 contains chips numbered 6, 11, 16, 21 and 26. Two chips, one from each box, are drawn at random. The numbers written on these chips are multiplied. The probability for the product to be an even number is (A) ⁄ (C) ⁄ ⁄ (B) ⁄ (D)

exponential distribution f(x) =

e ,x≥

The probability that the conversation will exceed five minutes is (A) e (C) (B) e (D) e IN - 2008 8. Consider a Gaussian distributed random variable with zero mean and standard deviation . The value of its cummulative distribution function at the origin will be (A) 0 (C) 1 (B) 0.5 (D) σ 9.

A random variable is uniformly distributed over the interval 2 to 10. Its variance will be ⁄ ⁄ (A) (C) (B) 6 (D) 36

IN - 2013 13. A continuous random variable X has probability density f(x) = . Then P(X > 1) is (A) 0.368 (C) 0.632 (B) 0.5 (D) 1.0 IN - 2014 14. Given that x is a random variable in the r nge , - with prob bility density function

the value of the constant k is

___________________ IN - 2009 10. A screening test is carried out to detect a certain disease. It is found that 12% of the positive reports and 15% of the negative reports are incorrect. Assuming that the probability of a person getting a positive report is 0.01, the probability that a person tested gets an incorrect report is (A) 0.0027 (C) 0.1497 (B) 0.0173 (D) 0.2100

15.

IN - 2010 11. The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05mm respectively. Assuming Gaussian distribution of

The figure shows the schematic of production process with machines A,B and C. An input job needs to be preprocessed either by A or by B before it is fed to C, from which the final finished product comes out. The probabilities of failure of the machines are given as:

Assuming independence of failures of the machines, the probability that a given job is successfully processed (up to third decimal place)is ______________

th

th

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GATE QUESTION BANK

Mathematics


4. [Ans. D] The number of ways coming 8 and 9 are (2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5), (5,4),(6,3) Total ways =9 So Probability of coming 8 and 9 are

[Ans. D]

5.

[Ans. D] X and Y are independent ∴ ( ) ( ) ( ) re true Only (D) is odd one

6.

[Ans. A] Number of favourable cases are given by HHHT HHTH HTHH THHH Total number of cases = 2C1 2C1 2C1 2C1 =16

So probability of not coming these

2.

[Ans. B] Probability of defective item = Probability of not defective item = 1 0.1 = 0.9 So, Probability that exactly 2 of the chosen items are defective = ( ) ( )

3.

[Ans. B]

∴ Probability = 7.

[Ans. A] A uniform function

Mean (t)̅ = ∫ t f(t) dt ∫ t( t 6

t)dt t

t 6

7

[

]

∫ t(

[

t

t)dt

t)dt

=∫ (t =0

t )dt 1

0

7

density

Density function

 1  f(x)   b  a  0

∫ t (

t)dt

a,x  b a  x,x  b

 Mean E(x)=

t)dt

b

x(F(x))  x a

ab 2

 Variance = F(x)2  f(x)

2

1

2

b   x F(x)  xF(x) x a  x a  b

= Standard deviation = √v ri nce =

and

0,x  a x  a  f(x)   f  x dx   , axb 0 b  a  xb 0,

]

∫ t (

distribution

x

Variance = ∫ t f(t)dt =∫ t (

( oth defective) S mple sp ce

( oth defective)

2

Put the value of F(x), we get √

2

1 1 b  dx  x. dx  Variance  x ba x a  x a b  a  b

th

th

2

th


GATE QUESTION BANK b

 x3   xL      3(b  a) a  2 b  a  

Mathematics 3

2

1 7  (3  3  1)    2 8



b3  a3 (b2  a2 )2  3(b  a) 4 b  a 2



(b  a)(b2  ab  a2 ) (b  a)2(b  a)2  2 3(b  a) 4 b  a



4b2  4ab  4a2  3a2  3b2  6ab 12



b2  a2  2ab 12

9.

[Ans. C] Probability of drawing 2 washers, first followed by 3 nuts, and subsequently the 4 bolts

10.

[Ans. D] Required probability =

(b  a)2  12

. / . /

Standard deviation = √v ri nce

(b  a)2 12 (b  a)  12



11.

[Ans. D] Given 4R and 6B , -

12.

[Ans. C]

Given: b=1, a=0

 Standard deviation =

8.

10 1  12 12

[Ans. D] Let probability of getting atleast one head = P(H) then P (at least one head) = 1 P(no head)  P(H)=1 P(all tails) But in all cases, 23=8

1 7 8 8

X=0

 P (H) = 1  

(X ) is Below X (X ) has to be less than 0.5 but greater than zero

Alternately Probability of getting at least one head ( ) ( )

13.

1 7 1   8 8 Alternately From Binomial theorem Probability of getting at least one head pq ( )

( )

X=1

[Ans. D] A event that he knows the correct answer B event that student answered correctly the question P(B) = ? ( )

(

)

( )

th

th

th


GATE QUESTION BANK

( )

he knows correct nswer

(

) (

( ⁄ )

14.

No. of employed men = 80% of men = 80 No. of employed women = 50% of women = 50 Probability if the selected one person being employed = probability of one employed women +probability of one employed man

⏟

⏟

e does not know correct nswer so he guesses

( ) ( ⁄ ) ) ⁄ ( ) ⁄

[Ans. D] x 1 2 P(x) 0.3 0.6 (x) (x) x

18.

V(x) x (

σ

(x )

[Ans. A]

3 0.1 So from figure Mean value = 1 V ri nce : μ me n x defective pieces (x μ) σ ) n(n ( ) ( ) ( ) ( )

(x) x (x) σ

Mathematics

, (x)-

(x) ( x (x)) ) ( )

√ ( )

15.

[Ans. A] 19.

16.

[Ans. *](Range 49 to 51)

[Ans. *] Range 0.25 to 0.27 p orm l distribution

q

Given that μ σ x μ x z σ ere x μ , s x gre ter th n z ) ence prob bility (z

Using Binomial distribution (x ≥ )

17.

( ) ( )

( ) ( )

-

∫ e dz σ√ ∴ of s ving ccount holder

[Ans. *] Range 0.64 to 0.66 Let number of men = 100 Number of women = 100 th

th

th


GATE QUESTION BANK

20.

[Ans. B] Mean m = np = 5.2 me (x ) e

25 Calculators

m

23 Non-defective

2 Defective

e

5 Calculators

e (x

Mathematics

) 4 Non-defective

1 Defective

CE 1.

2.

[Ans D] A, B, C are true (D) is not true. Since in a negatively skewed distribution mode > median > mean [Ans. D] Let the mean and standard deviation of the students of batch C be μ and σ respectively and the mean and standard deviation of entire class of first year students be μ and σ respectively Now given, μ σ and μ σ In order to normalise batch C to entire class, the normalize score must be equated since Z = Z =

=

Now Z =

p( defective in c lcul tors)

4.

[Ans. C] σ μ

5.

[Ans. B] Given f(x) = x for x = 0 else where (

)

∫ f(x)dx

∫ x dx

=0 1 The probability expressed in percentage P= = 2.469% = 2.47% 6.

[Ans. A] Given P(private car) = 0.45 P(bus 1 public transport) = 0.55 Since a person has a choice between private car and public transport P(public transport) = 1 – P(private car) = 1 – 0.45 =0.55 P(bus) = P(bus public transport) (bus public tr nsport) (public tr nsport) = 0.55 × 0.55 = 0.3025 ≃ 0.30 Now P(metro) = 1 [P(private car) + P(bus)] = 1 (0.45 + 0.30) = 0.25

=

Equation these two and solving, we get =  x = 8.969 ≃ 9.0 3.

x

[Ans. B] Since population is finite, hypergeometric distribution is applicable

th

th

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GATE QUESTION BANK

∴ P(private car) = 0.45 P(bus) = 0.30 and P(metro) = 0.25 7.

12.

[Ans. D] ere μ cm; σ ( x 102) =P.

[Ans. 6] ∫ f(x)dx ( x

f(x)

{

∴∫

( x

cm 6

x

Mathematics

)

x

)dx

x

x

x

(

)

x otherwise

x7

/ [

=P( x ) This area is shown below:

[

(

(

)]

]

[

-0.44

)

]

The shades area in above figure is given by F(0) –F ( 0.44) =

( )

(

(

)(

)

= 0.5 – 0.3345 = 1.1655 ≃ 16.55% Closest answer is 16.7% 8.

)

13.

[Ans. 0.4] (

)

∫ f( )d

[Ans. C] ( )|

P(2 heads) = 9.

[Ans. C] P(one ball is Red & another is blue) = P(first is Red and second is Blue)

14.

= 10.

[Ans. A] Given μ = 1000, σ = 200 We know that Z When X= 1200, Z Req. Prob = P (X (Z ) ( Z Less than 50%

11.

∫ d

[Ans. D] (X ) ( )

(X

)

(

[Ans. *] Range 0.26 to 0.27 Avg= 5 Let x denote penalty (x ) (x ) (x ) (x ) (x ) e ew (x n) x e e e ) p(x e

[

)

e

]

)

)

(X

15.

[Ans. B] S * T+ n( ) ( ) n(S)

16.

[Ans. *] Range 0.25 to 0.28 ( t) e (n t) n

)

( )

th

th

th


GATE QUESTION BANK

no of vehicles (

)

veh km

e

.

m ke ex ctly ‘ ’ moves nd ‘U’ moves in any order. Similarly to reach (10, 10) from (0,0) the robot h s to m ke ‘ ’ moves nd ‘U’ moves in any order. The number of ways this can be done is same as number of permutations of a word consisting of 10 ‘ ’ s nd ’U’s Applying formula of permutation with limited repetitions we get the answer as

/

= 2.e = 0.2707 CS 1.

[Ans. A] P: Event of selecting Box P, Q: Event of selecting Box P P(P)=1/3, P(Q)=2/3 P(R/P)=2/5, P(R/Q)=3/4

P(R/P).P(P) P(R/P).P(P)  P(R/Q)P(Q) 2/51/3   4/19 2/51/3  3/ 4  2/3 P(P/R)=

2.

5.

[Ans. D] The robot can reach (4,4) from (0,0) in 8C ways as argued in previous problem. 4 Now after reaching (4,4) robot is not allowed to go to (5,4) Let us count how many paths are there from (0,0) to (10,10) if robot goes from (4,4) to (5,4) and then we can subtract this from total number of ways to get the answer. Now there are 8C4 ways for robot to reach (4,4) from (0,0) and then robot takes the ‘U’ move from ( ) to ( ) ow from (5,4) to (10,10) the robot has to make 5 ‘U’ moves nd ‘ ’ moves in ny order which can be done in 11! ways = 11C5 ways Therefore, the number of ways robot can move from (0,0) (10,10) via (4,4) – (5,4) move is

[Ans. C] If f (x) is the continuous probability density function of a random variable X then, ( x b) P( x b) b

= f  x  dx

 a

3.

4.

[Ans. A] The probability that exactly n elements are chosen =The probability of getting n heads out of 2n tosses =

(

) . /

= =

(

) (

Mathematics

(Binomial formula) )

8C 4

[Ans. A] Consider the following diagram (3,3)

11C 5

 8   11   4 5 

=   

No. of ways robot can move from (0,0) to (10,10) without using (4,4) to (5,4) move is

 20  8   11        ways  10   4   5 

(0,0) The robot can move only right or up as defined in problem. Let us denote right move by ‘ ’ nd up move by ‘U’ ow to reach (3, 3), from (0,0) , the robot has to

which is choice (D)

th

th

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GATE QUESTION BANK

6.

[Ans. D] umber of permut tions with ‘ ’ in the first position =19! Number of permutations with ‘ ’ in the second position = 10 18! (Fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18! ways) umber of permut tions with ‘ ’ in rd position =10 9 17! (Fill the first 2 place with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers) nd so on until ‘ ’ is in th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers v il ble to fill before the ‘ ’ So the desired number of permutations which satisfies the given condition is

8.

[Ans. C] Let C denote computes science study and M denotes maths study. P(C on Monday and C on Wednesday) = P(C on Monday, M on Tuesday and C on Wednesday) + P(C on Monday, C on Tuesday and C on Wednesday) =1 0.6 0.4+ 1 0.4 0.4 = 0.24 + 0.16 = 0.40

9.

[Ans. B] It is given that P (odd) = 0.9 P (even) Now since 𝜮P(x) = 1 ∴ P (odd) + P (even) = 1 0.9 P (even) + P (even) = 1 P(even) =

…

∴ P(2) = P(4) = P(6) = P(even) )

=

Which are clearly not choices (A), (B) or (C) 7.

/ = P (z ≥

.z

/ = P (z ≥

(z

) = P (z ≥

(

)

)

10. _____(i)

)

(

)

(

)

P(f ce )

) (

) (

(0.5263)

= 0.1754 It is given that P(even | face > 3) = 0.75

[Ans. A] Given μ = 1, σ = 4 σ =2 and μ = 1, σ is unknown Given, P(X ) = P (Y ≥ 2 ) Converting into standard normal variates, .z

= 0.5263

Now, it is given that P(any even face) is same i.e. P(2) = P(4) = P(6) Now since, P(even) = P(2) or P(4) or P(6) = P(2) + P(4) + P(6)

… Now the probability of this happening is given by = (

Mathematics

= 0.75

= 0.75 ( )

)

( )

=1

decl red f ulty

f ulty

p

q p

σ =3

not f ulty

decl red not f ulty decl red f ulty

q

q

decl red not f ulty

From above tree (decl red f ulty) th

=0.468

[Ans. A] The tree diagram of probabilities is shown below q

Now since we know that in standard normal distribution P (z ) = P (z ≥ 1) _____(ii) Comparing (i) and (ii) we can say that

=

th

pq th

(

q)(

p)


GATE QUESTION BANK

11.

[Ans. A] If b c … Then, no. of divisors of (x )(y )(z )… iven ∴ o of ivisors of ( )( ( )( )

(

(

which are multiples

13.

14.

15.

[Ans. C] (x ) , (x)V(x) Where V(x) is the variance of x, Since variance is σ and hence never negative, ≥

( t le st one he d)

TT )

So required prob bility

)

∴ equired rob bility 12.

)

( ∪ )

16. No. of divisors of of o of divisors of ( )( )

Mathematics

[Ans. B] Required Probability = P (getting 6 in the first time) + P (getting 1 in the first time and getting 5 or 6 in the second time) + P (getting 2 in the first time and getting 4 or 5 or 6 in the second time) + P (getting 3 in the first time and getting 3 or 4 or 5 or 6 in the second time) ( )

( )

( )

17.

[Ans. C] The p.d.f of the random variable is x +1 P(x) 0.5 0.5 The cumulative distribution function F(x) is the probability upto x as given below x +1 F(x) 0.5 1.0 So correct option is (C)

18.

[Ans. C] e (k)

[Ans. A] + The five cards are * Sample space ordered pairs st nd P (1 card = 2 card + 1) )( )( )( )+ *(

k P is no. of cars per minute travelling.

[Ans. D] 𝛔y = a 𝛔x is the correct expression Since variance of constant is zero.

For no cars. (i.e. k = 0) For no cars. P(0) e So P can be either 0,1,2. (i.e. k = 0,1,2) For k = 1, p(1)=e

[Ans. A] Let A be the event of head in one coin. B be the event of head in second coin. The required probability is * ) ( ∪ )+ ( )| ∪ ) ( ∪ ) ( ) ( ∪ ) ( ) (both coin he ds)

For k = 2 , P(2)= Hence ( ) e e th

( )

( )

e 4

th

e 5

th


GATE QUESTION BANK

e (

(

20.

(

)

∴ equired prob bility is ( ∪ ( )

∪ )

⁄ )

(

( ) (

)

e

[Ans. *] Range 0.24 to 0.27 The smaller sticks, therefore, will range in length from almost 0 meters upto a maximum of 0.5 meters, with each length equally possible. Thus, the average length will be about 0.25 meters, or about a quarter of the stick.

24.

(

)

[Ans. 10] 22 occurs in following ways 6 6 6 4 w ys 6 6 5 5 w ys

[Ans. 0.25] ( ∪ ) P(S) = 1 ( ) ( ) ( ) utu lly exclusive ( ) ( ) ( ) et ( ) x; ( ) x P(A) P(B) = x( x) Maximum value of y = x ( x) dy ( x) x dx = 2x = 1 x

equired prob bility

(max)

x 21.

( )

) e

19.

( )

Mathematics

ximum v lue of y

[Ans. *] Range 11.85 to 11.95 For functioning 3 need to be working (function)

ECE 1.

(

)

[Ans. D]

3 1  6 2 3 1 P(even number )   6 2 Since events are independent, therefore 1 1 1  P(odd/even)    2 2 4 P(Odd number) 

p 22.

[Ans. *] Range 3.8 to 3.9 Expected length = Average length of all words

2.

[Ans. A] I

∫ e

(

√ omp ring with (

23.

σ√ ut μ

[Ans. *] Range 0.259 to 0.261 Let A = divisible by 2, B = divisible by 3 and C = divisible by 5, then n(A) = 50, n(B) = 33, n(C) = 20 n( ) n( ) n( ) n( ) P( ∪ ∪ )

∫

∫ rom x σ σ th

)

th

)

e

dx

σ

√

dx

e

dx

…

nd x

th


)

GATE QUESTION BANK

Put σ ∫

3.

in

Probability of failing in paper 2, P (B) = 0.2 Probability of failing in paper 1, when

equ tion

e

√

A   0.6 B  A  P A  B We know that, P   PB  B

student has failed in paper 2, P

[Ans. C] 

 P x.dx  1





 Ke



Mathematics

(

∴

.dx  1

ax

)

e dx

∫

e

dx

 x x,for x  0      x for x  0 K K   1 a a 

( )

= 0.6 0.2 = 0.12



or ∫

( )

7.

[Ans. A] CDF: F x  

x

 PDF

dx



For x<0, F x  

x

  x  1

dx

1

4.

[Ans. D] . / ( )

P (Y/D) =

. / ( )

. / ( )

= 5.

. / ( )

=0.4

0

1

F0 

1 2

conc ve upw rds

For x>0, F x   F0 

x

 x  1

dx

0

[Ans. A] var[x]=σ =E[(x x)2] Where, x=E[x] x= expected or mean value of X defining

 1  x2    x  concave downwards 2  2  Hence the CDF is, shown in the figure (A).



E[X] =

 xf  xdx x



8.

[Ans. A]

Given: Px  x   Me2|x|  Ne3|x|



    x P xi   x  xi dx    i



 P  xdx  1 x

 xiP xi 

 



i

Variance σ is a measure of the spread of the values of X from its mean x. Using relation , E[X+Y]= E[X]+E[Y] And E[CX]=CE[X] On var[x]= σ =E[(x x)2] σ = ,Xx2 = E[X2] [ ,X-] 6.











Me2|x|  Ne3|x| dx  2 Me2|x|  Ne3|x| dx  1 0

By simplifying

2 3

 M N 1 9.

[Ans. B] x+y=2 x y=0 => x =1, y = 1 P(x=1,y=1) = ¼

[Ans. C] Probability of failing in paper 1, P (A) = 0.3 th

th

¼ = 1/16 th


GATE QUESTION BANK

10.

[Ans. C]

14.

Probability of getting head in first toss = Probability of getting head in second toss =

[Ans. C] P(no. of tosses is odd) (no of tosses is

…)

P(no. of toss is 1) = P(Head in 1st toss)

and in all other 8 tosses there should

P(no. of toss is 3) = P(tail in first toss, tail in second toss and head in third toss)

be tail always. So probability of getting head in first two tosses ( )( )( )…………… ( ) = (1/2)10

P(no. of toss is 5) = P(T,T,T,TH) . /

11.

Mathematics

[Ans. B] Both the teacher and student are wrong Mean = ∑ k = 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3.0 E(x2) = ∑ k

… etc

So, P(no. of tosses in odd)

⁄ ⁄ ⁄ ⁄

Variance(x)= E(x2) – * (x)+ =10.2 9=1.2 12.

[Ans. D] P(H, H, H, T) +P (H, H, H, H ) =

13.

. /

. /

15.

[Ans. B] ( V ≥ V) ( V V≥ ) *z v v+ Linear combination of Gaussian random variable is Gaussian ∴ (z ≥ ) and not mean till zero because both random variables has mean zero hence ( ) Hence Option B is correct

16.

[Ans. D] F(x) = P{X x} (x) * X x+ x 2X 3

. / =

[Ans. C] Total number of cases = 36 Favorable cases: (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) Total number of favorable cases

For positive value of x, (x) (x) is always greater than zero For negative value of x (x) (x)is ve ut , (x) (x)- x ≥

Then probability th

th

th


GATE QUESTION BANK

17.

18.

[Ans. *] Range 0.65 to 0.68 et ‘ ’ be different types of f milies nd ‘S’ be there siblings. S S S (siblings) Probability that child chosen at random having sibling is 2/3

(x)

et S

,

∑x f(x)

[Ans. C]

21.

[Ans. *] Range 2.9 to 3.1 Let the first toss be head. Let x denotes the number of tosses(after getting first head) to get first tail. We can summarize the even as Event(After x Probability getting first H) T 1 1/2 HT 2 1/2 1/2=1/4 HHT 3 1/8 nd so on…

II)gives

(

)S

……

(x) i.e. The expected number of tosses (after first head) to get first tail is 2 and same can be applicable if first toss results in tail. Hence the average number of tosses is

22.

-

20.

(I

(II)

S

∑x …

……

(I)

S

f(x) ∴ (x)

……

S

[Ans. *] Range 0.32 to 0.34 This is a tricky question, here, X X X independent and identically distributed random variables with the uniform distribution , -. So, they are equiprobable. So X X or X have chances being largest are equiprobable. So, [P {X is largest}] or [P {X is largest}] or [P {X is largest}] =1 and P {X is largest} = P {X is largest} = P {X is largest}

[Ans. *] Range 49.9 to 50.1 Set of positive odd number less than 100 is 50. As it is a uniform distribution

∑ x (x) …

∴ *X is l rgest + 19.

Mathematics

[Ans. *] Range 0.15 to 0.18 X X X X X X et z X X X (X ) X X (z ) Pdf of z we need to determine. It is the convolution of three pdf

(z

23.

th

)

∫

z

dz

z

|

[Ans. *] Range 0.79 to 0.81 |x| ,|x|- ∫ e dx √

th

th


GATE QUESTION BANK

√ √ √ √ √

∫ |x| exp 4

x x

∫ x exp 4

∴pr(P ∪ Q) pr(P) + pr(Q) (D) is true since P Q P n(P Q) n(P) pr(P Q) pr(P)

5 dx 5 dx

5 dx 2.

x

∫ x exp 4

x

∫ x exp 4

√ [ exp (

√ , 24.

x

∫ |x| exp 4

-

[Ans. B] P(A|B) =

5 dx 5 dx

x ) dx]

( he d in tosses nd first toss in he d) = P(HHT) + P(HTH)

4/5

Parcel is sent to R

∴ Required probability = R

3.

1/5

S

Parcel is lost Parcel is lost

parcel

is

=

[Ans. C] If two fair dices are rolles the probability distribution of r where r is the sum of the numbers on each die is given by r P(r)

4/5

that

) ( )

Also, P(first toss is head) =

√

Parcel is sent to

Probability

(

∴ P(2 heads in 3 tosses | first toss is head) ( he ds in tosses nd first toss in he d) (first toss is he d)

[Ans. *] Range 0.43 to 0.45 Pre flow diagram is

1/5

Mathematics

lost

2 Probability that parcel is lost by 3 Probability that parcel is lost by provided that the parcel is lost

4 5

EE 1.

6 [Ans. D] (A) is false since of P & Q are independent pr(P Q) = pr(P) pr(Q) which need not be zero. (B) is false since pr(P ∪ Q) = pr(P) + pr(Q) – pr(P Q) (C) is false since independence and mutually exclusion are unrelated properties.

7 8 9 10 11 12 th

th

th


GATE QUESTION BANK

The above table has been obtained by taking all different all different ways of obtaining a particular sum. For example, a sum of 5 can be obtained by (1, 4), (2, 3), (3, 2) and (4, 1). ∴ P(x = 5) = 4/36 Now let us consider choice (A) Pr(r > 6) = Pr(r ≥ 7)

P(1, 5, 6) =

=

P(3, 4, 5) =

=

P(1, 2, 5) =

=

∴ Choice (C) P(1, 5 and 6) = 5.

is correct.

[Ans. C] x is uniformly distributes in [0, 1] ∴ Probability density function

= =

Mathematics

=

f(x) =

∴ Choice (A) i.e. pr(r > 6) = 1/6 is wrong. Consider choice (B) pr(r/3 is an integer) = pr(r = 3) + pr (r = 6) + pr (r = 9) + pr (r = 1)

=

=1

∴ f(x) = 1 0 < x < 1 Now E(x ) = ∫ x f(x)dx ∫x

dx

= =

6.

=

[Ans. B] Let N people in room. So no. of events that at least two people in room born at same date

∴ Choice (B) i.e. pr (r/3) is an integer = 5/6 is wrong. Consider choice (C) Now, pr(r/4 is an integer) = pr(r = 4) + pr (r = 8) + pr (r = 12) = =

≥

N

Solving, we get N = 7

+

7.

[Ans. C] (II is red|I is white) (II is red nd I is white) (I is white) (I is white nd II is red) (I is white)

8.

[Ans. B]

=

pr(r = 8) = ∴ pr(r = 8 | r/4 is an integer) =

…

=

= ∴ Choice (C) is correct. 4.

[Ans. C] Dice value 1 2

Probability

and

is the entrie

3

rectangle The region in which maximum of {x, y} is

4

less than ⁄

is shown below as shaded

region inside this rectangle

5 6

th

th

th


GATE QUESTION BANK y (

)

(

12.

)

Mathematics

[Ans. *] Range 0.35 to 0.45 (

)

x ∫

dx

∫

x|

(

(

)

dx

∫

dx

x|

)

13. p .m x,x y-

∫ f(x)dx

[Ans. *] Range 0.4 to 0.5 ∫ f(x) dx

/

by property

∴ ∫ kx dx k 9.

14.

[Ans. A] (x

) ,e

10.

∫ e dx e -

, e -

e

[Ans. B] Let number of heads = x, ∴ Number of tails n x ∴ ifference x (n x)or (n x n or n x If x n n x n x

n

If n

x

n

IN 1.

x

x

|

[Ans. D]

=52 weeks and 2 days are extra. Out of x)

x

7, (SUN MON) (or) (SAT SUN) are favorable. So, Probability of this event= 2.

or x

[Ans. C] Since the reading taken by the instrument is normally distributed, hence P(x

x )

Where, [Ans. *] Range 0.13 to 0.15 Let proportionality constant = k ∴ ( dot) k ( dots) k ( dots) k ( dots) k ( dots) k ( dots) k ∴k k k k k k k ∴k ∴ rob bility of showing dots

∴k

[Ans. D] Since leap chosen will be random, so, we assume it being the case of uniform probability distribution function. Number of days in a leap year=366 days

As x and n are integers, this is not possible ∴ Probability 0 11.

k

Now

√

√

∫ e

(

)

.dx

μ e n of the distribution σ St nd rd devi tion of the distribution. ∫

exp(

)dx

where, n=x 10 (∴μ kg) and from the data given in question √

∫

e

dx

On equating, we get 0.05=0.84 σ k

σ

th

th

th


GATE QUESTION BANK

3.

[Ans. D] Mean=

8.

[Ans. B] By definition of Gaussian distribution, total area under the curve =1. Hence half of the area =0.5

9.

[Ans. A]

=5.9 V. (

√

S

̅)

(

̅)

(

̅)

V (closest answer is 0.2)

P(x)= 4.

[Ans. C] ( )

=

Mean = μ ( )

∫ x (x)dx = ∫

Var(x)= ∫ (x

1 2  3 3 5.

Mathematics

=∫

[Ans. A] ]

(x

(x)dx

μ)

) dx =

10.

[Ans. C] Probability of incorrect report

11.

[Ans. C] σ mm μ mm Then probability



 P(x)dx  1

x dx = 6



 Area under triangle =

c 1 2

α 6.

[Ans. A] Probability that the sum of digits of two dices is even is same either both dices shows even numbers or odd numbers on the top of the surface ( ) ∴ ( ) ( ) Where ( ) Probability of occurring even number of both the dices ( ) Probability of occurring odd number of both the dices (

(X where x

(

X

μ

σ mm

(

)

)

( )

e

√ e

√

So, number of measurement more than 10.15mm P Total number of measurement

)

nd (

)

) ≃

∴ ( ) 12. 7.

[Ans. A] ∫ f(x) dx=P or ∫

e

or e

|

.dx =P

[Ans. D] For the product to be even, the numbers from both the boxes should not turn out to be odd simultaneously. ∴ ( )

( )( )

or P = .

th

th

th


GATE QUESTION BANK

13.

[Ans. A] ∫ f(x)dx e |

14.

15.

∫ e dx

e

[Ans. 2] For valid pdf ∫ ∫

Mathematics

dx

pdf dx

;

;k

[Ans. *] Range 0.890 to 0.899 Probability that job is successfully processed = ( )( )

th

th

th


GATE QUESTION BANK

Mathematics

Numerical Methods ME – 2005 1. Starting from x = 1, one step of Newton – Raphson method in solving the equation x³ +3x 7=0 gives the next value (x₁) as (A) x₁=0.5 (C) x₁ = .5 (B) x₁= . 0 (D) x₁=2 2.

With a 1 unit change in b, what is the change in x in the solution of the system of equation = 2 .0 0. = (A) Zero (C) 50 units (B) 2 units (D) 100 units

ME – 2006 3. Match the items in columns I and II. Column I Column II (P) Gauss-Seidel (1) Interpolation method (Q) Forward (2) Non-linear Newton-Gauss differential method equations (R) Runge-Kutta (3) Numerical method integration (S) Trapezoidal (4) Linear algebraic Rule equation (A) 2 (B) 2 (C) 2 (D) 2 4.

Equation of the line normal to function ) f(x) = (x (A) y = x 5 (B) y = x 5

at (0 5) is (C) y = x (D) y = x

5 5

ME – 2007 5. A calculator has accuracy up to 8 digits 2

after decimal place. The value of

 sinxdx 0

when evaluated using this calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is (A) 0.00000 (C) 0.00500 (B) 1.0000 (D) 0.00025

ME – 2010 6. Torque exerted on a flywheel over a cycle is listed in the table. Flywheel energy (in J per unit cycle) using impson’s rule is Angle (degree) Torque (N-m) 0 0 60 1066 120 323 180 0 240 323 300 55 360 0 (A) 542 (C) 1444 (B) 992.7 (D) 1986 ME – 2011 7.

The integral ∫

dx, when evaluated by

using impson’s / rule on two equal subintervals each of length 1, equals (A) 1.000 (C) 1.111 (B) 1.098 (D) 1.120 ME – 2013 8. Match the correct pairs. Numerical Order of Fitting Integration Scheme Polynomial . impson’s / 1. First Rule Q. Trapezoidal Rule 2. Second . impson’s / 3. Third Rule (A) P – 2 , Q – 1, R – 3 (B) P – 3, Q – 2 , R – 1 (C) P – 1, Q – 2 , R – 3 (D) P – 3, Q – 1 , R – 2 ME – 2014 9. Using the trapezoidal rule, and dividing the interval of integration into three equal sub intervals, the definite integral ∫ |x|dx is ____________

th

th

th


GATE QUESTION BANK

10.

The value of ∫ .

( )

“value approximate estimate?

calculated using

the Trapezoidal rule with five sub intervals is _______ 11.

The definite integral ∫

12.

The real root of the equation 5x 2cosx = 0 (up to two decimal accuracy) is _______

13.

Consider

an

equation

= t

.If x =x at t = 0 , the

CE – 2005 Linked Answer Question 1 and 2 Give a>0, we wish to calculate its reciprocal value 1/a by using Newton Raphson Method for f(x) = 0.

2.

The Newton Raphson algorithm for the function will be (A) x

= (x

)

(B) x

= (x

x )

(C) x

= 2x

ax

(D) x

=x

x

in

the

(C) – (D)

CE – 2007 4. The following equation needs to be numerically solved using the NewtonRaphson method x3 + 4x – 9 = 0 the iterative equation for the purpose is (k indicates the iteration level)

differential

increment in x calculated using RungeKutta fourth order multi-step method with a step size of Δt = 0.2 is (A) 0.22 (C) 0.66 (B) 0.44 (D) 0.88

1.

(A) – (B) 0

value”)

is evaluated

using Trapezoidal rule with a step size of 1. The correct answer is _______

ordinary

Mathematics

For a = 7 and starting with x = 0.2 the first two iteration will be (A) 0.11, 0.1299 (C) 0.12, 0.1416 (B) 0.12, 0.1392 (D) 0.13, 0.1428

CE – 2006 3. A 2nd degree polynomial f(x) has values of 1, 4 and 15 at x = 0, 1 and 2 respectively.

5.

(A) x

=

(B) x

=

(C) x

=x

(D) x

=

x

Given that one root of the equation x 10x + 31x – 30 = 0 is 5, the other two roots are (A) 2 and (C) and (B) 2 and (D) 2 and

CE – 2008 6. Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. Given ∑x = 6, ∑y = 2 ∑x = and ∑xy = the values of a and b are respectively (A) 2 and 3 (C) 2 and 1 (B) 1 and 2 (D) 3 and 2 CE – 2009 7. In the solution of the following set of linear equation by Gauss elimination using partial pivoting 5x + y + 2z = 34; 4y – 3z = 12; and 10x – 2y + z = 4; the pivots for elimination of x and y are (A) 10 and 4 (C) 5 and 4 (B) 10 and 2 (D) 5 and 4

The integral ∫ f(x) dx is to be estimated by applying the trapezoidal rule to this data. What is the error (define as true th

th

th


GATE QUESTION BANK

CE – 2010 8. The table below given values of a function F(x) obtained for values of x at intervals of 0.25. x 0 0.25 0.5 0.75 1.0 F(x) 1 0.9412 0.8 0.64 0.50 The value of the integral of the function between the limits 0 to using impson’s rule is (A) 0.7854 (C) 3.1416 (B) 2.3562 (D) 7.5000 CE 9.

2011 The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x = 0. If i denotes the iteration index, the correct iteration scheme will be (A) x (B) x

= (x

)

= (x

CE – 2013 12. Find the magnitude of the error (correct to two decimal places) in the estimation of following integral using impson’s ⁄ Rule. Take the step length as 1.___________ ∫ (x

1.

Consider

)

(D) x

= (x

)

he error in

 xe dx

=

)

(

)

is 2

1 R xn1   xn   can be used to compute 2 xn 

.

the (A) square of R (B) reciprocal of R (C) square root of R (D) logarithm of R

0 .

CE – 2012 The estimate of ∫ .

1 3

to an accuracy of at least 106

The Newton-Raphson iteration

for a continuous

The values of and ( ) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h = 0.02 is approximately (A) . 0 (C) .5 0 (B) .0 0 (D) .0 0

11.

x

1

function estimated with h=0.03 using the central difference formula f(x)|

=

using the trapezoidal rule is (A) 1000e (C) 100e (B) 1000 (D) 100

f(x)|

(

series

CS – 2008 2. The minimum number of equal length subintervals needed to approximate

3. 10.

the

= 0.5 obtained from the NewtonRaphson method. The series converges to (A) 1.5 (C) 1.6 (D) 1.4 (B) √2

)

= (x

0) dx

CS – 2007

2

(C) x

Mathematics

obtained using

impson’s rule with three – point function evaluation exceeds the exact value by (A) 0.235 (C) 0.024 (B) 0.068 (D) 0.012

CS – 2010 4. Newton-Raphson method is used to compute a root of the equation x 13 = 0 with 3.5 as the initial value. The approximation after one iteration is (A) 3.575 (C) 3.667 (B) 3.677 (D) 3.607

th

th

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GATE QUESTION BANK

CS – 2012 5. The bisection method is applied to compute a zero of the function f(x) = x x x in the interval [1,9]. The method converges to a solution after ___________ iterations. (A) 1 (C) 5 (B) 3 (D) 7 CS – 2013 6. Function f is known at the following points: x f(x) 0 0 0.3 0.09 0.6 0.36 0.9 0.81 1.2 1.44 1.5 2.25 1.8 3.24 2.1 4.41 2.4 5.76 2.7 7.29 3.0 9.00 he value of ∫ f(x)dx computed using the trapezpidal rule is (A) 8.983 (C) 9.017 (B) 9.003 (D) 9.045 CS – 2014 7. The function f(x) = x sin x satisfied the following equation: ( ) + f(x) + t cos x = 0. The value of t is _________. 8.

In the Newton-Raphson method, an initial guess of = 2 made and the sequence x x x .. is obtained for the function 0.75x 2x 2x =0 Consider the statements (I) x = 0. (II) The method converges to a solution in a finite number of iterations. Which of the following is TRUE? (A) Only I

Mathematics

(B) Only II (C) Both I and II (D) Neither I nor II 9.

With respect to the numerical evaluation of the definite integral,

= ∫ x dx where a and b are given, which of the following statements is/are TRUE? (I) The value of K obtained using the trapezoidal rule is always greater then or equal to the exact value of the defined integral (II) The value of K obtained using the impson’s rule is always equal to the exact value of the definite integral (A) I only (B) II only (C) Both I and II (D) Neither I nor II ECE– 2005 1. Match the following and choose the correct combination Group I Group II (A) Newton1. Solving nonRaphson linear equations method (B) Runge-Kutta 2. Solving linear method simultaneous equations (C) impson’s 3. Solving ordinary Rule differential equations (D) Gauss 4. Numerical elimination integration method 5. Interpolation 6. Calculation of Eigen values (A) A-6, B-1, C-5, D-3 (B) A-1, B-6, C-4, D-3 (C) A-1, B-3, C-4, D-2 (D) A-5, B-3, C-4, D-1

th

th

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GATE QUESTION BANK

ECE– 2007 2. The equation x3 x2+4x 4=0 is to be solved using the Newton-Raphson method. If x=2 is taken as the initial approximation of the solution, then the next approximation using this method will be (A) 2/3 (C) 1 (B) 4/3 (D) 3/2

(A) 2

x

x

.

sin x ..

2 cos x

x

.

..

(C) 2

x

x

.

..

(D) 2

x

x

.

..

8.

The series ∑

eXn 1  eXn X2  eXn 1  Xn   1 (D) Xn1  n Xn -eXn

ECE– 2014 6. The Taylor expansion of is

x

Match the application to appropriate numerical method. Application Numerical Method P1:Numerical M1:Newtonintegration Raphson Method P2:Solution to a M2:Runge-Kutta transcendental Method equation P3:Solution to a M : impson’s system of linear 1/3-rule equations P4:Solution to a M4:Gauss differential equation Elimination Method (A) P1—M3, P2—M2, P3—M4, P4—M1 (B) P1—M3, P2—M1, P3—M4, P4—M2 (C) P1—M4, P2—M1, P3—M3, P4—M2 (D) P1—M2, P2—M1, P3—M3, P4—M4

(C) Xn1  1  Xn 

ECE– 2013 5. A polynomial f(x) = a x a x a x a x a with all coefficients positive has (A) No real root (B) No negative real root (C) Odd number of real roots (D) At least one positive and one negative real root

(B) 2

7.

ECE– 2008 3. The recursion relation to solve x= using Newton-Raphson method is (A) =e (B) = e

ECE– 2011 4. A numerical solution of the equation f(x) = x √x = 0 can be obtained using Newton – Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (A) 0.306 (C) 1.694 (B) 0.739 (D) 2.306

Mathematics

converges to

(A) 2 ln 2 (B) √2

(C) 2 (D) e

EE– 2007 1.

The differential equation

=

is

discretised using Euler’s numerical integration method with a time step T > 0. What is the maximum permissible value of T to ensure stability of the solution of the corresponding discrete time equation? (A) 1 (C) (B) /2 (D) 2 EE– 2008 2. Equation e = 0 is required to be solved using ewton’s method with an initial guess x = . Then, after one

th

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GATE QUESTION BANK

step of ewton’s method estimate x of the solution will be given by (A) 0.71828 (C) 0.20587 (B) 0.36784 (D) 0.00000 3.

A differential equation dx/dt = e u(t) has to be solved using trapezoidal rule of integration with a step size h = 0.01 sec. Function u(t) indicates a unit step function. If x(0)= 0, then value of x at t = 0.01 s will be given by (A) 0.00099 (C) 0.0099 (B) 0.00495 (D) 0.0198

EE– 2009 4. Let x 7 = 0. The iterative steps for the solution using Newton – aphson’s method is given by (A) x

= (x

(B) x

=x

(C) x

=x

(D) x

=x

)

EE– 2013 7. When the Newton – Raphson method is applied to solve the equation f(x) = x 2x = 0 the solution at the end of the first iteration with the initial guess value as x = .2 is (A) 0.82 (C) 0.705 (B) 0.49 (D) 1.69 EE– 2014 8. The function ( ) = is to be solved using Newton-Raphson method. If the initial value of is taken as 1.0, then the absolute error observed at 2nd iteration is ___________ IN– 2006 1. For k = 0 2 . the steps of Newton-Raphson method for solving a non-linear equation is given as

2 5 xk 1  xk  xK2 . 3 3 (x

Starting from a suitable initial choice as k tends to , the iterate tends to (A) 1.7099 (C) 3.1251 (B) 2.2361 (D) 5.0000

)

EE– 2011 5. Solution of the variables and for the following equations is to be obtained by employing the Newton-Raphson iterative method equation(i) 0x inx 0. = 0 equation(ii) 0x 0x cosx 0. = 0 Assuming the initial values = 0.0 and = .0 the jacobian matrix is 0 0. 0 0. (A) * (C) * + + 0 0. 0 0. 0 0 0 0 (B) * (D) * + + 0 0 0 0 6.

Mathematics

IN– 2007 2. Identify the Newton-Raphson iteration scheme for finding the square root of 2.

3.

(A) x

=

(x

)

(B) x

= (x

)

(C) x

=

(D) x

= √2

x

The polynomial p(x) = x + x + 2 has (A) all real roots (B) 3 real and 2 complex roots (C) 1 real and 4 complex roots (D) all complex roots

Roots of the algebraic equation x x x = 0 are ) (A) ( (C) (0 0 0) (B) ( j j) (D) ( j j)

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GATE QUESTION BANK

IN– 2008 4. It is known that two roots of the nonlinear equation x3 – 6x2 +11x 6 = 0 are 1 and 3. The third root will be (A) j (C) 2 (B) j (D) 4

IN– 2013 8. While numerically solving the differential equation


=

with

x(0) = 0 and the constant 0 is to be numerically integrated using the forward Euler method with a constant integration time step T. The maximum value of T such that the numerical solution of x converges is (C) (A) (B)

2xy = 0 y(0) =

using

Euler’s predictor – corrector (improved Euler – Cauchy )method with a step size of 0.2, the value of y after the first step is (A) 1.00 (C) 0.97 (B) 1.03 (D) 0.96

IN – 2009 5.

Mathematics

IN– 2014 9. The iteration step in order to solve for the cube roots of a given number N using the Newton- aphson’s method is

(D) 2

(A) x

=x

(B) x

= (2x

(C) x

=x

(D) x

= (2x

(

x ) )

(

x ) )

IN– 2010 6. The velocity v (in m/s) of a moving mass, starting from rest, is given as

=v

t.

Using Euler’s forward difference method (also known as Cauchy-Euler method) with a step size of 0.1s, the velocity at 0.2s evaluates to (A) 0.01 m/s (C) 0.2 m/s (B) 0.1 m/s (D) 1 m/s IN– 2011 7. The extremum (minimum or maximum) point of a function f(x) is to be determined by solving

( )

= 0 using the

Newton-Raphson method. Let f(x) = x x and x = 1 be the initial guess of x. The value of x after two iterations (x ) is (A) 0.0141 (C) 1.4167 (B) 1.4142 (D) 1.5000

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GATE QUESTION BANK

Mathematics


y = sin ( ) = 2

[Ans. C] By N-R method ,

=x –

f(x) = x  f( ) =

x

x =x

(

)

(

)

( ) ( )

y = sin (

x =

y = sin( ) = 0 5 y = sin ( ) =

7

y = sin (

 f (x) = x f ( )= , 

=1

(

)

) = 0.70 0

)=

7 y = sin ( ) =

[Ans. C] Given x y = 2 (i) .0 x 0.0 y = b (ii) Multiply 0.99 is equation (i) and subtract from equation (ii); we get ( .0 0. )x = b (2 0. ) 0.02x = b . 0.02Δx = Δb Δx =

0.02

[Ans. D]

4.

[Ans. B] Given f(x) = (x 2 ) f (x) = (x

f(x)dx = [(y

∫ y ∫

0.70 0

6.

y )

[(0

0)

0.70 0

0.70 0

[(0

y )]

7.

[Ans. C] x y= ∫

( 0

0)

2( 2.7 /unit cycle.

=

Slope of normal = 3 (∵ roduct of slopes = 1) Slope of normal at point (0, 5) y 5 = (x 0) y= x 5 [Ans. A] b a 2 0 h= = = n y = sin(0) = 0

0=0

[Ans. B] ower = ω = Area under the curve. h (y = [(y y ) y y )

/

Slope of tangent at point (0, 5) 2 ) / = m = (0

y

2(0.70 0

2(y

)

2(y

)]

sinx dx =

=

5.

)=0

Trapezoidal rule

= 50 units

3.

0.70 0

(0.5) = .5

=

y = sin ( 2.

0.70 0

1 1

x dx = (y

= (

55)

2 )]

3

2 h

x

2

2

0

y 2

y )

)

= . 8.

[Ans. D] By the definition only

y = sin ( ) = 0.70 0

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GATE QUESTION BANK

9.

[Ans. *] Range 1.10 to 1.12

By intermediate value theorem roots lie be between 0 and 1. et x = rad = 57. 2 By Newton Raphson method f(x ) x =x f (x ) 2x sin x 2 cos x x = 5 2 sin x x = 0.5 2 x = 0.5 25 x = 0.5 2

∫ |x|dx is h ∫ ydx = [y 2

2(y

y

y

x

1

y

1

0.33

2

y ]

y

0.33

2

∫ |x|dx =

.)

y

[

y

0.33

1

0.333

1

2(0.

0.

)]

13.

= . 0 10.

[Ans. *] Range 1.74 to 1.76 2.5 h= = 0. 5 y 2y 2y ∫ . ln (x)dx = [ 2y y =

.

[ln(2.5)

2ln( . ) = .75 11.

2(ln2. )

.

2y

]

2 ln( . )

CE 1.

[Ans. *]Range 1.1 to 1.2

t|

Δx = 0.0

0. = 0.

= 2t

t|

Set up the equation as x = i.e. = a

∫ f(x)dx = [y

y

2(y

iven in question 0 1 1 2 1 0.5

∫ dx = [y x 2

.

.

t Δx = 2

To calculate using N-R method

rapezoidal rule

x y

)dt

[Ans. C]

∫ dx by trapezoidal rule x

h=

[Ans. D] The variation in options are much, so it can be solved by integrating directly dx = t dt ∫ dx = ∫ ( t

ln( )]

2ln( .7)

Mathematics

y

y

..y

)]

a=0 i.e. f(x) =

2 3 0.33

a=0

Now f (x) = f(x ) =

a

f (x ) =

2(y )]

For N-R method = [ 2 = .

0.

2

0.5]

x

=x x

12.

[Ans. *] Range 0.53 to 0.56 Let f(x) = 5x 2 cos x f (x) = 5 2 sin x f(0) = f( ) = 2.

=x

(

)

(

) (

)

Simplifying which we get x = 2x ax

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GATE QUESTION BANK

2.

[Ans. B] For a = 7 iteration equation Becomes x = 2x 7x with x = 0.2 x = 2x 7x = 2× 0.2 – 7(0.2) = 0.12 and x = 2x 7x = 2× 0.12 7(0. 2) = 0.1392

x

[Ans. A] f(x) = 1, 4, 15 at x = 0, 1 and 2 respectively ∫ f(x)dx = (f

2f

=x x

=

5.

f )

∫ f(x)dx = (1 + 2(4) + 15) = 12

and α β

4.

αβ

(

βγ =

)

= 30

βγ

γα = 5β

βγ

5γ =

=

 5 (β γ) βγ = ince βγ = from (i) 5 (β γ) = β γ=5 βγ = olving for β and γ β (5 β) = β 5β =0  β = 2 and γ = Alternative method 5 1 0 31 0 0 5 25 30 1 5 6 0 (x 5)(x )=0 5x (x 5)(x 2)(x )=0 x=2 5

x )dx + =

2=

[Ans. A ] Given f(x) = x x =0 f (x) = x Newton – Raphson formula is

γα =

 βγ = (i) Also

Error = exact – Approximate value =

βγ

α βγ = 5

Now exact value ∫ f(x)dx

= *x

2x x

αβγ=

Approximate value by rapezoidal ule = 12 Since f(x) is second degree polynomial, let f(x) = a0 + a x + a x f(0) = 1 a 0 0= a = f(1) = 4 a a a =  1+ a a = a a =  f(2) = 15 a 2a a = 5  2a a = 5  2a a = Solving (i) and (ii) a = and a = f(x) = 1 – x + 4 x x

=

[Ans. A] Given x – 10 x + 31x 30 = 0 One root = 5 Let the roots be α β and γ of equation ax + bx + cx + d = 0

(3 points Trapezoidal Rule) Here h = 1

=∫ (

f(x ) f (x ) (x x ) ( x ) x x x ( x )

=x

x 3.

Mathematics

6.

[Ans. D] Y = a + bx Given n= ∑x = and ∑xy = th

th

∑y = 2 ∑x = 14

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GATE QUESTION BANK

Normal equations are ∑y = na b∑x ∑xy = a∑x b∑x Substitute the values and simply a= b=2 7.

9.

5 [0 0

2

(

)

| 2] → 2

0 [0 5

2

=

f(x ) f(x )

=x

=x

(

x

)

2x

x

=

2x

2

[x

x

]

10.

[Ans. D] Error in central difference formula is (h) This means, error If error for h = 0.03 is 2 0 then Error for h = 0.02 is approximately (0.02) 2 0 0 (0.0 )

11.

[Ans. D] Exact value of ∫ .

| 2] 2

So the pivot for eliminating x is a = 10 Now we eliminate x using this pivot as follows : 0 2 [0 | 2] 5 2 5 0 2 0 2] → [0 0 2 /2 Now to eliminate y, we need to compare the elements in second column at and below the diagonal element Since a = 4 is already larger in absolute value compares to a = 2 The pivot element for eliminating y is a = 4 itself. The pivots for eliminating x and y are respectively 10 and 4 8.

[Ans. A] x

[Ans. A] The equation is 5x + y + 2z = 34 0x + 4y – 3z = 12 and 10x – 2y + z = The augmented matrix for gauss elimination is 5 2 [0 | 2] 0 2 Since in the first column maximum element in absolute value is 10 we need to exchange row 1 with row 3

Mathematics

.

dx = .0

Using impson’s rule in three – point form, b a .5 0.5 h= = = 0.5 n 2 So, x 0.5 1 1.5 y 2 1 0.67 ∫

= =

[

0.5

] [2

0. 7

]

= . So, the estimate exceeds the exact value by Approximate value – Exact value = 1.1116 1.0986 =0.012(approximately)

12.

[Ans. *](Range 0.52 to 0.55) Using impson’s ule X 0 1 2 3 Y 10 11 26 91

4 266

[Ans. A] I = h(f = 0. = 0.7 5

f

2f

f

0.25(

0.

0.

0.5)

∫ (x

f ) 2

= [( 0

2

0)dx 2

)

2(2 )

(

)]

= 2 5. The value of integral th

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GATE QUESTION BANK

∫ (x

0) dx = *

=

0

=2

x 5

=

0x+ 3.

.

x

=

[Ans. A] +

, x = 0.5

)

(α

)

2α =

=x =α

2α = α + R α =R α=√ So this iteration will compute the square root of R

α= + α= 8α = 4α +9 α = 4.

α = = 1.5 [Ans. A] Here, the function being integrated is f(x) = xe f (x) = xe + e = e (x + 1) f’ (x) = xe + e + e = e (x + 2) Since, both are increasing functions of x, maximum value of f ( ) in interval 1 2, occurs at = 2 so ( )| (2 max |f =e 2) = e Truncation Error for trapezoidal rule = TE (bound)

[Ans. D] y=x dy = 2x dx f(x)= x x

= .5

5.

=

(b – a) max |f ( )| 1

=

(2 – 1) [e (2 + 2)]

=

e

Now putting

(

)

=

= 57 7

)=5 f(x ) 2 oot lies between and

x =(

0

0

)=2 f(x ) 0 2 After ' ' interations we get the root

x =(

= max |f ( )|

.

[Ans. B] f( ) = 5 f( ) = 5 72 ) ) f( 0 f( 0

is number of subintervals

=

.

= . 07

max |f ( )|

Where

(x

2α=α+ =

when the series converges x

=

= 1000 e

At convergence x =x =α α=

Given x

2.

)/

[Ans. C]

5 Magnitude of error = 2 5. 2 . = 0.5 CS 1.

(

Mathematics

2

6.

[Ans. D] h ∫ f(x)dx = [f(0) 2 =

0

=

=

0 [

. .

[

0.

f( )

2(0.0 0. . . 7.2 )

2(∑f )] ]

5 . ]

= 9.045

h= Now, No. of intervals,

= th

th

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GATE QUESTION BANK

7.

8.

9.

ECE 1.

2.

[Ans. – ] Given (x) + f(x) + t cos x = 0 and f(x) = x sin x f (x) = x cos x + sin x f (x) = x ( sin x) + cos x + cos x = 2 cos x – x sin x = 2 cos x – f (x) 2 cos x – f (x) + f(x) +t cos x = 0 2 cos x = tcos x t = 2 [Ans. A] f(x) = 0.75x 2x 2x f (x) = 2.25x x 2 x =2 f = 2 f = f x =x =0 f f = f = 2 f x =x =2 f f = 2 f = f x =x =0 f Also, root does not lies between 0 and 1 So, the method diverges if x = 2 nly ( )is true.

x1  2  3.

 1  x n  4.

x1  x0 

e e

e xn 1  exn

[Ans. C] x

f(x ) f (x )

=x

f(2) = (2

x

) = √2

√2

f (2) =

√

=2

and

√

=

√ )

(√ √

= .

√

5.

[Ans. D] f(x) = a x a x a x a x a If the above equation have complex roots, then they must be in complex conjugate pair, because it’s given all co-efficients are positive ( they are real ) So if complex roots are even no. (in pair) then real roots will also be even. ption ( )is wrong From the equation ( 0) roduct of roots = As no. of roots = 4, Product of roots < 1 either one root 0 (or) Product of three roots < 0 ption ( )is rong. Now, take option (A), Let us take it is correct . Roots are in complex conjugate pairs = Product of roots 0 | | | | 0 which is not possible ption (A) is wrong orrect answer is option ( )

[Ans. C] By definition (& the application) of various methods

4=0

Next approximation x1  x0 

8 4  12 3

[Ans. C] Given : f(x)= x e By Newton Raphson method, f(x ) x x =x =x f (x )

[Ans. C] For value of K if trapezoidal rule is used then the value is either greater than actual value of definite integral and if impson’s rule is used then value is exact Hence both statements are TRUE

[Ans. B] y(t) =x3 x2 + 4x x0 = 2

Mathematics

f  x0  f ' x0 

x03  x02  4x0  4 3x02  2x0  4 th

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GATE QUESTION BANK

6.

[Ans. A] sin x 2 cos x x = (x – )

7.

[Ans. B]

8.

[Ans. D] ∑

n

2( –

x 2

.

x

)

Put x = as given, x = [e ( 2) ]/e = 0.71828

)

[Ans. C]

=e

.. = e

u(t)

x

x 2

.

. . x in t

[Ans. D] Here,

x = ∫ e u(t) dt = ∫ f(t) dt At t = 0.01, x = Area of trapezoidal

= x

f(x y) =

x

h

=(

)x h

4.

x

5.

=[

=*

( )

f(x ) = e f’(x ) = e

6.

(

]

)

(

)

+

0x cos x 0x sinx

20x

0sinx ] 0cosx

0 0

is

0 + 0

[Ans. D] x x x =0 (x )(x )=0 x =0 x =0 x= x= j

=x –(

(

The matrix at x = 0 x =

( )

=

.

[Ans. B] u(x x ) = 0x sin x 0. = 0 v(x x ) = 0x 0x cosx 0. = 0 The Jacobian matrix is u u x x v v [ x x ]

[Ans. A] Here f(x) = e f (x) = e The Newton Raphson iterative equation is

=

=x

= *x

|

Δ 2 o maximum permissible value of Δ is 2 .

i.e. x

e

[Ans. A]

since h = Δ here Δ

x

[

= x

h

=x

.

= 0.0099

h

or stability |

x

f(0.0 )] =

= [f(0)

Euler’s method equation is x = x h. f(x y ) x x = x h( )

2.

i=0

(

x =

2

as e = EE 1.

Now put

3.

=

Mathematics

)

)

th

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GATE QUESTION BANK

7.

[Ans. C] x =x =

.2

Hence, it will have atleast 5 (0+1)= 4 complex roots.

f(x ) f (x ) ( .2) 2( .2) ( .2) 2

4.

[Ans. C] Approach- 1 Given, x3 – 6x2 + 11x – 6 = 0 Or (x – 1)(x – 3)(x – 2) = 0 x= 1, 2, 3.

= 0.705 8.

[Ans. *] Range 0.05 to 0.07 Clearly, x = 0 is root of the equation f(x) = e =0 f (x) = e and x = .0 Using ewton raphson method (e . ) f(x ) x =x = = f (x ) e. e and x = x

f(x ) = f (x ) e =

(e

Approach- 2 For ax3 +bx2 + cx +d = 0 If the three roots are p,q,r then Sum of the roots= p+q+r= b/a Product of the roots= pqr= d/a pq+qr+rp=c/a

) e

5.

[Ans. D] dx x = dt f(x, y) =

e

e = 0. 7 0. = 0.0 Absolute error at 2nd itteration is |0 0.0 | = 0.0 IN 1.

x

=x h

=( [Ans. A] As k ∞ xk+1 ≈xk xk = x

h (x y ) = x )x

2.

3.

h

h(

x

)

)

h

|

h

x

/

(

or stability |

Δ

x = x x =5 x =5

Mathematics

Δ

= 1.70

[Ans. A] Assume x = √ f(x) = x =0 f(x ) x =x = [x f (x ) 2

6.

[Ans. A] dv =v t dt t v dv =v t dt 0 0 0 0+0 0. = 0 0.1 0 0+0.1 0. = 0.0

7.

[Ans. C] f(x) = x x f (x) = x = g(x) x = initial guess g (x) = x g (x ) x =x g (x )

2 ] x

[Ans. C] Given p(x) = x + x + 2 There is no sign change, hence at most 0 positive root ( rom escarte’s rule of signs) p( x) = x x+2 There is one sign change, hence at most 1 negative root ( rom escarte’s rule of signs)

2

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GATE QUESTION BANK

(

= x =x = .5 = .

)

= .5

g(x ) g (x ) 0.75 7

8.

[Ans. D] dy = 2xy x = 0 y = h = 0.2 dx y =y h. f(x y ) (0.2)f(0 ) = = and y = y [f(x y ) f(x y )] (0. )[f(0 ) f(0.2 )] = = 0. is the value of y after first step, using Euler’s predictor – corrector method

9.

[Ans. B] For convergence x

Mathematics

= x =x x=

x =

(2x

x

)

x= √

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GATE QUESTION BANK

Mathematics

Calculus ME – 2005 1.

ME – 2006

The line integral ∫ ⃗ ⃗⃗⃗⃗ of the vector function ⃗ ( ) 2xyz ̂+ x²z + ̂ k²y ̂ from the origin to the point P (1,1,1) (A) is 1 (B) is Zero (C) is 1 (D) cannot be determined without specifying the path

2.

be (A) (B) 8.

(A)

(C)

(B)

(D)

Changing the order of the integration in the double integral I = ∫ ∫

(

∫ (

What is q?

(A)

(C) X (D) 8 )

(A)

√

(C)

√

(B)

√

(D)

.

(

10.

/

)

(A) 0 (B) ⁄

is equal to 11.

∫

(C) (D) 1

The area of a triangle formed by the tips of vectors a , b and c is (A)

(

)(

(D) Zero

(B)

|(

)

Stoke’ theorem connects (A) A line integral and a surface integral (B) Surface integral and a volume integral (C) A line integral and a volume integral (D) Gradient of function and its surface integral

(C)

|

(D)

(

(C) 2∫ (

√

ME – 2007

(B) 2∫

6.

1 and t is a real number,

Let x denote a real number. Find out the INCORRECT statement + represents the set if all (A) S * real numbers greater then 3 + represents the empty (B) S * set + represents the (C) S * union of set A and set B + represents the set (D) S * of all real umbers between a and b, where and b are real number

)

leads to (A) 4y (B) 16y²

(C) 0 (D) ⁄

dt is:

∫

9.

4.

)

⁄ ⁄

√

By a change of variables x(u,v) = uv, y(u,v) = v/u is double integral, the integral f(x,y) changes to f(uv, u/v) ( ). Then, ( ) (A) 2 v/u (C) v² (B) 2 u v (D) 1

I =∫ ∫ (

2x2  7x  3 , then limf(x) will x 3 5x2  12x  9

Assuming i =

The right circular cone of largest volume that can be enclosed by a sphere of 1 m radius has a height of

3.

5.

If f( x ) =

7.

)

th

th

) (

)|

| )

th


GATE QUESTION BANK

12.

√

If

√

y (2) = (A) 4 or 1 (B) 4 only 13.

, then

√

14.

20.

The divergence of the vector field )̂ ( ) ̂ is (x y) ̂ ( (A) 0 (C) 2 (B) 1 (D) 3

21.

Let

(C) 1 (D) 1/ln2

y2  4x and x2  4y is ⁄ (A) (B) 8 23.

⁄ (C) (D) 16

The distance between the origin and the point nearest to it on the surface

The directional derivative of the scalar

z2  1  xy is

function f(x, y, z) = x2  2y2  z at the

(A) 1

point P = (1, 1, 2) in the direction of the vector ⃗ ̂ ̂ is (A) 4 (C) 1 (B) 2 (D) 1

at x=2, y=1?

ME – 2009 22. The area enclosed between the curves

Which of the following integrals is unbounded? (C) ∫ (A) ∫ (D) ∫

What is

(A) 0 (B) ln2

The length of the curve

(B) ∫ 16.

In the Taylor series expansion of ex about x = 2, the coefficient of (x 2)4 is ⁄ (A) ⁄ (C) ⁄ (B) (D) ⁄

The minimum value of function y = x2 in the interval [1, 5] is (A) 0 (C) 25 (B) 1 (D) Undefined

between x = 0 and x = 1 is (A) 0.27 (C) 1 (B) 0.67 (D) 1.22 15.

19.

(C) 1 only (D) Undefined

ME – 2008

Mathematics

(C) √ (D) 2

(B) √ ⁄ 24.

A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of  x  y  on path AB 2

17.

Consider the shaded triangular region P shown in the figure. What is∬ xydxdy? y

P

18.

B

X

x

A

2

⁄ ⁄

The value of (A) (B)

Y

x+2y=2

1

0 (A) (B)

traversed in a counter-clockwise sense is

⁄ ⁄

(C) ⁄ (D) 1

(

)

(C) (D)

25.

is ⁄ ⁄

th

(A)

(C)

(B)

(D) 1

The divergence of the vector field ̂ at a point (1,1,1) is ̂ ̂ equal to (A) 7 (C) 3 (B) 4 (D) 0 th

th


GATE QUESTION BANK

ME – 2010 26. Velocity vector of a flow field is given as ⃗ ̂ .̂ The vorticity vector at (1, 1, 1) is (A) 4 ̂ ̂ (B) 4 ̂ ̂ 27.

The function (A) o o

(C) ̂ (D) ̂

∀ ∀ (B) o o ∀ ∀ except at x = 3/2 (C) o o ∀ ∀ except at x = 2/3 (D) o o ∀

28.

29.

ME – 2012 33. Consider the function ( ) in the interval . At the point x = 0, f(x) is (A) Continuous and differentiable. (B) Non – continuous and differentiable. (C) Continuous and non – differentiable. (D) Neither continuous nor differentiable.

̂ ̂

R R R R

34.

R R

ME – 2011 30. If f(x) is an even function and is a positive real number, then ∫ ( )dx equals

31.

What is (A) (B)

32.

36.

For the spherical surface the unit outward normal vector at the point

is

(C) π (D) π

(C)

(B) (C)

is

√

(A) (B)

has

/

√

̂

√

̂

√

̂

√

̂

√

(C) ̂ (D) 37.

equal to?

A series expansion for the function (A)

.

∫ ( )

(C) 0 (D) 1

(C) 1 (D) 2

At x = 0, the function f(x) = (A) A maximum value (B) A minimum value (C) A singularity (D) A point of inflection

R except at x = 3 ∀ R

(D)

/ is

35.

The parabolic arc is √ revolved around the x-axis. The volume of the solid of revolution is (A) π (C) π (B) π (D) π

(A) 0 (B)

. (A) 1/4 (B) 1/2

The value of the integral ∫ (A) –π (B) –π

Mathematics

√

̂

√

̂

√

̂

The area enclosed between the straight line y = x and the parabola y = in the x – y plane is (A) 1/6 (C) 1/3 (B) 1/4 (D) 1/2

ME – 2013 38. The following surface integral is to be evaluated over a sphere for the given steady velocity vector field defined with respect to a cartesian coordinate system having i, j and k as unit base vectors. ∫∫ (

(D) th

th

) th


GATE QUESTION BANK

Where S is the sphere, and n is the outward unit normal vector to the sphere. The value of the surface integral is (A) π (C) π⁄ (B) π (D) π 39.

45.

If a function is continuous at a point, (A) the limit of the function may not exist at the point (B) the function must be derivable at the point (C) the limit of the function at the point tends to infinity (D) the limit must exist at the point and the value of limit should be same as the value of the function at that point

The value of the definite integral ( )

∫ √

is

(A)

√

(C)

√

(B)

√

(D)

√

46.

Divergence of the vector field ̂ ( ̂ ̂ ) is (A) 0 (C) 5 (B) 3 (D) 6

(C) 3 (D)Not defined

47.

The value of the integral

ME – 2014 40.

is (A) 0 (B) 1

∫ 41.

Which one of the following describes the relationship among the three vectors ̂ ̂ ̂ ̂ + ̂ + ̂ ̂ ̂ ̂ (A) The vectors are mutually perpendicular (B) The vectors are linearly dependent (C) The vectors are linearly independent (D) The vectors are unit vectors

42.

.

(

)

/ is equal to

(A) 0 (B) 0.5 43.

̂

̂

)̂ )̂ ̂ ̂

̂ ̂ ̂ ̂

)

(

)

) (

)

(A) 3 (B) 0 48.

(C) 1 (D) 2

The value of the integral ∫ ∫ is (

)

(C) (

(B) (

)

(D) .

(A)

) /

).

Where, c is the square cut from the first quadrant by the lines x = 1 and y = 1 will ( G ’ h o o h h line integral into double integral) (A) ⁄ (C) ⁄ (B) 1 (D) ⁄

̂ ̂

( (

CE – 2005 1. Value of the integral ∮ (

(C) 1 (D) 2

Curl of vector ⃗ ̂ (A) ( (B) ( (C) (D)

44.

Mathematics

̂ ̂ 2.

A rail engine accelerates from its stationary position for 8 seconds and travels a distance of 280 m. According to the Mean Value theorem, the speedometer at a certain time during acceleration must read exactly. (A) 0 kmph (C) 75 kmph (B) 8 kmph (D) 126 kmph

The best approximation of the minimum value attained by (100x) for ≥ is _______

th

th

th


GATE QUESTION BANK

CE – 2006 3. What is the area common to the circles o 2 (A) 0.524 a (C) 1.014 a2 (B) 0.614 a2 (D) 1.228 a2 4.

The directional derivative of f(x, y, z) = 2 + 3 + at the point P (2, 1, 3) in the direction of the vector a= k is (A) (C) (B) (D)

CE – 2007 5. Potential function  is given as = . When will be the stream function () with the condition  = 0 at x = y = 0? (A) 2xy (C) (B) + (D) 2 6.

Evaluate ∫ (C) π⁄ (D) π⁄

(A) π (B) π⁄ 7.

10.

12.

transformed to (A) (B)

9.

(C) √ (D) 18

parabola is y = 4h

(A) ∫ √ √

√

(D)

√

√

(C) ∫

= 0 by substituting (C)

where x is the

horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is

= 0 can be

(D)

14.

√

∫

.

The



/

is

(A) 2/3 (B) 1

The inner (dot) product of two vectors ⃗ and ⃗ is zero. The angle (degrees) between in two vectors is (A) 0 (C) 90 (B) 30 (D) 120

(C) 40.5 (D) 54.0

√

(B) 2∫

+

is

CE – 2010 13. A parabolic cable is held between two supports at the same level. The horizontal span between the supports is L. The sag at the mid-span is h. The equation of the

CE – 2008 +

)

For a scalar function f(x, y, z) = the directional derivative at the point P(1, 2, 1) in the ⃗ is direction of a vector (A) (B)

A velocity is given as ̅ = 5xy + 2 y2 + 3yz2⃗ . The divergence

The equation

The value of ∫ ∫ ( (A) 13.5 (B) 27.0

CE – 2009 11. For a scalar function f(x, y, z) = + 3 + 2 the gradient at the point P (1, 2, 1) is (A) 2 + 6 + 4⃗ (C) 2 + 12 + 4⃗ (D) √ (B) 2 + 12 – 4⃗

of this velocity vector at (1 1 1) is (A) 9 (C) 14 (B) 10 (D) 15

8.

Mathematics

15.

th

(C) 3/2 (D)

Given a function ( ) The optimal value of f(x, y) th

th


GATE QUESTION BANK

(A) Is a minimum equal to 10/3 (B) Is a maximum equal to 10/3 (C) Is a minimum equal to 8/3 (D) Is a maximum equal to 8/3

CE – 2013 21.

CE – 2011 16.

∫

√ √

√

?

22.

(C) a (D) 2a

/

o

magnitudes a and b respectively. |⃗ ⃗ | will be equal to (A) – (⃗ ⃗ ) (C) + (⃗ ⃗ ) (B) ab ⃗ ⃗ (D) ab + ⃗ ⃗ CE – 2012 19. For the parallelogram OPQR shown in the sketch, ̅̅̅̅ ̂ ̂ and ̅̅̅̅ R ̂ .̂ The area of the parallelogram is Q

(C) 1 (D)

24.

A particle moves along a curve whose parametric equation are : and z = 2 sin (5t), where x, y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm ) at t = 0 is ___________

25.

If {x} is a continuous, real valued random variable defined over the interval ( ) and its occurrence is defined by the density function given as:

(C) 1 (D) π

If ⃗ and ⃗ are two arbitrary vectors with

R P

.

( )

/

√

wh

‘ ’

‘ ’

the statistical attributes of the random variable {x}. The value of the integral

O

(A) ad –bc (B) ac+bd

.

With reference to the conventional Cartesian (x, y) coordinate system, the vertices of a triangle have the following coordinates: ( ) ( ) ( ) ( ) ( ) ( ). The area of the triangle is equal to (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

π

(A) 0 (B) π

(C) ad + bc (D) ab – cd

∫

o

o

. √

/

dx is

(A) 1 (B) 0.5

The infinity series

(A) sec (B)

(C) 1 (D) 8/3

23.

π

20.

o

(A) (B)

Wh ho h o λ such that the function defined below is continuous π ? f(x)={

18.

The value of ∫ (A) 0 (B) 1/15

CE – 2014

(A) 0 (B) a/2 17.

Mathematics

o

26.

(C) o (D)

(C) π (D) π⁄

The expression

o

(A) log x (B) 0 th

th

(C) x log x (D) th


GATE QUESTION BANK

CS – 2005 1.

Let G(x) 

CS – 2010

1   g(i)xi where |x|<1. 2 (1  x) i0 

What is g(i)? (A) i (B) i+1

(C) 2i (D) 2i

CS – 2007 2. Consider the following two statements about the function f(x) =|x|: P: f(x) is continuous for all real values of x Q: f(x) is differentiable for all real values of x Which of the following is true? (A) P is true Q is false (B) P is false Q is true (C) Both P and Q are true (D) Both P and Q are false CS – 2008 3.

4.

x sinx equals Lim x   x cosx (A) 1 (B) 1

(C) (D)

Let P=∑

∑

7.

What is the value of (A) 0 (B)

∫ (A) 0 (B) 1

(C) (D) 1

(A) 0 (B) 2

(C) –i (D) i

CS – 2012 9. Consider the function f(x)= sin(x) in the interval x ,π⁄ π⁄ -. The number and location(s) of the local minima of this function are (A) One , at π⁄ (B) One , at π⁄ (C) Two , at π⁄ and π (D) Two , at π⁄ and π CS – 2013 10. Which one the following function is continuous at x =3? (A) ( )

{

(B) ( )

2

(C) ( )

2

(D) ( ) CS – 2014 11. Let the function ( )

CS – 2009 6.

/ ?

∫

A point on a curve is said to be extreme if it is a local minimum or a local maximum. The number of distinct extrema for the 4 3 2 curve 3x  16x  24x  37 is (A) 0 (C) 2 (B) 1 (D) 3

.

CS – 2011 8. Given i = √ , what will be the evaluation of the definite integral

where k is a positive integer. Then (A) (C) (B) (D) 5.

Mathematics

(π (π

|

Where evaluates to

o o (π o (π

) )

) )

(π (π

)| )

1 and ( ) denote the

0

derivation of f with respect to . Which of the following statements is/are TRUE? (I) There exists

(C) ln2 (D) ln 2

. th

th

/

h h th

( )


GATE QUESTION BANK

(II) There exists . (A) (B) (C) (D) 12.

13.

14.

/

(A) h h

( )

I only II only Both I and II Neither I nor II

(B)

A function f (x) is continuous in the interval [0, 2]. It is known that f(0) = f(2) = 1 and f(x) = 1. Which one of the following statements must be true? (A) There exists a y in the interval (0,1) such that f(y) = f(y + 1) (B) For every y in the interval (0, 1), f(y) = f(2 y) (C) The maximum value of the function in the interval (0, 2) is 1 (D) There exists a y in the interval (0, 1) such that f(y) ( )

(C)

(D)

ECE – 2006 2. As x is increased from function f  x  

If and are 4 – dimensional subspace of a 6 – dimensional vector space V, then the smallest possible dimension of is ____________. If ∫

dx = π, then the value of k

e 1  ex



3 The integral sin  d is given by

3.

0

1 2 2 (B ) 3

o

(A) π (B) π

, the



The value of the integral given below is ∫

to

x

(A) monotonically increases (B) monotonically decreases (C) increases to a maximum value and then decreases (D) decreases to a minimum value and then increases

is equal to_______. 15.

Mathematics

4 3 8 (D) 3

(A) (C) – π (D) π

ECE – 2005 1. The derivative of the symmetric function drawn in given figure will look like

(C)

  P  ds , where P is a vector, is

4.

equal to (A) ∮ P  dl

(C) ∮  P  dl

(B) ∮  P dl

(D)

  Pdv

 P , where P is a vector, is equal to

5.

th

2 (A) P P  P

(C) 2P  P

2 (B)  P P

(D)  P 2P

th

th


GATE QUESTION BANK

ECE – 2007 6. For the function , the linear approximation around = 2 is (A) (3 x) (B) 1 x





ECE – 2008 12. Consider points P and Q in the x –y plane, with P=(1,0) and Q=(0,1). The line Q

integral 2  xdx  ydy  along the

 P

(C) 3  2 2  1  2 x e





2

semicircle with the line segment PQ as its diameter (A) is 1 (B) is 0 (C) is 1 (D) depends on the direction (clockwise or anticlockwise) of the semicircle

(D) 7.

8.

9.

10.

For x <<1, coth(x) can be approximated as (A) x (C) (B) x2 (D)

In the Taylor series expansion of exp(x)+sin(x) about the point x=π the coefficient of (x π)2 is (π) (A) (π) (C) (π) (π) (B) (D)

14.

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x=0? (A) sin(x3) (C) cos(x3) 2 (B) sin(x ) (D) cos(x2)

15.

The value of the integral of the function g(x, y)=4x3+10y4 along the straight line segment from the point (0,0) to the point (1,2) in the x y plane is (A) 33 (C) 40 (B) 35 (D) 56

16.

For real values of x, the minimum value of the function f(x)=exp(x)+ exp( x) is (A) 2 (C) 0.5 (B) 1 (D) 00

17.

Consider points P and Q in the x-y plane, with P=(1, 0) and Q= (0, 1).

Which one of the following function is strictly bounded? 2 (A) (C) x x (B) e (D)

(A) 0.5 (B) 1 11.

13. Consider the function f(x) = x – 2. The maximum value of f(x) in the closed interval [ 4,4] is (A) 18 (C) 2.25 (B) 10 (D) Indeterminate

sin   /2 lim is 0  (C) 2 (D) not defined

The following Plot shows a function y which varies linearly with x. The value of 2



the integral I  ydx is 1

Y 3 2

) along ∫ ( the semicircle with the line segment PQ as its diameter (A) Is (B) Is 0 (C) Is 1 h

1 1

Mathematics

1

(A) 1.0 (B) 2.5

2

3

X

(C) 4.0 (D) 5.0

th

th

th


GATE QUESTION BANK

(D) Depends on the direction (clockwise or anti-clockwise) of the semicircle

21.

ECE – 2009 18. The Taylor series expansion of

sinx at x   is given by x  (A) 1 

 x  2  ..... 3!

(B)

2 x    1   .....

(C)

2 x    1  .....

the value of the integral ∯ (A) 3V (B) 5V

3!

 x  

2

19.

3!

 .....

If a vector field ⃗ is related to another ⃗ , which vector field ⃗ through ⃗ = of the following is true? Note: C and refer to any closed contour and any surface whose boundary is C. (A) ∮ ⃗ ⃗ = ∬ ⃗ ⃗ (B) ∮ ⃗ ⃗ =

∬⃗ ⃗

(C) ∮

⃗ ⃗ =

∬⃗ ⃗

(D) ∮

⃗ ⃗ =

∬⃗ ⃗

ECE – 2010 20. If ⃗

̂ ̂ , then ∮ ⃗ ⃗⃗⃗ over the path shown in the figure is

(A) 0 (B) ⁄√

̂⃗

is

(C) 10V (D) 15V

ECE\IN – 2012 23. The direction of vector A is radially outward from the origin, with where and K is constant. The value of n for which . A = 0 is (A) 2 (C) 1 (B) 2 (D) 0 ECE\EE – 2012 24. The maximum value of ( ) in the interval [1,6] is (A) 21 (C) 41 (B) 25 (D) 46 ECE – 2013 25. The maximum value of unit which the approximation holds to within 10% error is (A) (C) (B) (D) 26.

√

, then has a maximum at minimum at maximum at minimum at

ECE – 2011 22. Consider a closed surface S surrounding a volume V. If is the position vector of a point inside S, with ̂ the unit normal of S,

3!

(D) 1 

If (A) (B) (C) (D)

Mathematics

The divergence of the vector field ⃗ ̂ ̂ ̂ is (A) 0 (B) 1/3

√

(C) 1 (D) 3

(C) 1 (D) 2√

th

th

th


GATE QUESTION BANK

27.

Consider a vector field ⃗ ( ) The closed loop line integral ∮ ⃗  can be expressed as ⃗ )  over the closed (A) ∯( surface boundary by the loop (B) ∰(  ⃗ )dv over the closed volume bounded by the loop (C) ∭(  ⃗ )dv over the open volume bounded by the loop ⃗ ) over the closed surface (D) ∬( bounded by the loop

Mathematics

34.

The magnitude of the gradient for the function ( ) at the point (1,1,1) is_______.

35.

The directional derivative of ( ) ( ) ( )in the direction √

of the unit vector at an angle of with y axis, is given by ________________. EE – 2005 1.

For the scalar field u =

, magnitude

of the gradient at the point (1, 3) is ECE – 2014 28. The volume under the surface z(x, y) = x+y and above the triangle in the x-y plane defined by {0 y x and 0 x 12} is______ 29.

30.

For function ( ) (A) o (B) o The value of

the maximum value of the occurs at (C) (D) o .

(A) (B)

(B) √ ⁄ 2.

For the function f(x) = , the maximum occurs when x is equal to (A) 2 (C) 0 (B) 1 (D) 1

3.

If S = ∫

/ is

31.

The maximum value of the function ( ) ( ) (wh ) occurs at x =____.

32.

The maximum value of ( ) 0 x 3 is ______.

in the interval

EE – 2006 4. A surface S(x, y) = 2x + 5y – 3 is integrated once over a path consisting of the points that satisfy (x+1)2+ (y 1)2 =√ . The integral evaluates to (A) 17√ (C) √ /17 (D) 0 (B) 17/√ 5.

33.

, then S has the value (C) ⁄ (D) 1

⁄ ⁄

(A) (B)

(C) (D)

(C) √ (D) ⁄

⁄

(A) √

The expression V = ∫ πR (

h

)

h

for the volume of a cone is equal to

For a right angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is (A) 12 (C) 60 (B) 36 (D) 45

th

(A) ∫ πR (

h

)

(B) ∫ πR (

h

)

(C) ∫

π

(

(D) ∫

π

(

th

h

R) h ⁄R )

th


GATE QUESTION BANK

EE – 2007 6.

EE – 2010

The integral equals (A) (B) 0

(

∫

o

) o

EE – 2009 8. f(x, y) is a continuous function defined over (x, y) [0, 1] [0, 1]. Given the two constraints, x > and y > , the volume under f(x, y) is (A) ∫

∫

(B) ∫

∫

(C) ∫

∫

9.

10.

√

11.

∫

√

(

)

(

)

( √

a minimum a discontinuity a point of inflection a maximum

Divergence of the three-dimensional radial vector field is (A) 3 (C) ̂ ̂ ̂ (B) 1/r (D) ̂ ( ̂ ̂)

13.

The value of the quantity P, where ∫ (A) 0 (B) 1

, is equal to (C) e (D) 1/e

EE – 2011 14. The two vectors [1, 1, 1] and [1, a, where a = . (A) (B) (C) (D)

)

A cubic polynomial with real coefficients (A) can possibly have no extrema and no zero crossings (B) may have up to three extrema and upto 2 zero crossings (C) cannot have more than two extrema and more than three zero crossings (D) will always have an equal number of extrema and zero crossings

has

12.

) (

At t = 0, the function ( ) (A) (B) (C) (D)

(C) (1/2) o (D) (1/2)

EE – 2008 7. Consider function f(x)= ( ) where x is a real number. Then the function has (A) only one minimum (B) only two minima (C) three minima (D) three maxima

(D) ∫

Mathematics

15.

√

],

/, are

orthonormal orthogonal parallel collinear

The function f(x) = 2x – has (A) a maxima at x = 1 and a minima at x=5 (B) a maxima at x = 1 and a minima at x= 5 (C) only a maxima at x = 1 (D) only a minima at x = 1

EE – 2013 16. Given a vector field , the line integral

F(x, y) = ( )̂ ( )̂ ’ line integral over the straight line from ( ) = (0, 2) to ( ) = (2, 0) evaluates to (A) –8 (C) 8 (B) 4 (D) 0

evaluated along a segment on the x∫ axis from x = 1 to x = 2 is (A) 2.33 (C) 2.33 (B) 0 (D) 7 17.

th

The curl of the gradient of the scalar field defined by (A) th

th


GATE QUESTION BANK

(B) (C) ( ( (D)

)

(

19.

20.

21.

23.

( ) Where f and v are scalar and vector fields respectively. If h is (A) (B) (C) (D)

24.

The minimum value of the function ( ) 0 in the interval , - is (A) 20 (C) 16 (B) 28 (D) 32

)

)

EE – 2014 18. Let ( ) . The maximum value of the function in the interval ( ) is (A) (C) (B) (D) The line integral of function , in the counterclockwise direction, along the circle is (A) π (C) π (B) π (D) π Minimum of the real valued function ( ) occurs at x equal to ( ) (A) (C) 1 (B) (D)

IN – 2005 1. A scalar field is given by f = x2/3 + y2/3, where x and y are the Cartesian coordinates. The derivative of f along the line y = x directed away from the origin, at the point (8, 8) is

To evaluate the double integral ∫ .∫

( ⁄ )

.

/

substitution u = (

/ dy, we make the ) and

2.

)

( ) ∫ (∫

)

( ) ∫ (∫

)

( ) ∫ (∫

)

(A)

√

(B)

√

(A) (B) 0 3.

√

∫ ( )

is

(C) f(1) (D) f(0)

The value of the integral ∫ (A) 2 (B) does not exist

(C) (D)

is 2

̅(t) has a constant magnitude, If a vector R then

4.

A particle, starting from origin at t = 0 s, is traveling along x-axis with velocity π π o . /

(D)

√

f(t) defined over [0,1],

̅ (A) R 22.

(C)

Given a real-valued continuous function

. The

integral will reduce to ( ) ∫ (∫

Mathematics

̅ (B) R 5.

̅

̅

̅ R ̅ (C) R ̅

̅ (D) R

̅ R

̅

If f = + …… + where ai (i = 0 to n) are constants,

At t = 3 s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is _________

then

is

(A) ⁄ (B) ⁄

th

th

(C) nf (D) n√

th


GATE QUESTION BANK

6.

The plot of a function f(x) is shown in the following figure. A possible expression for the function f(x) is f(x)

(B) once differentiable but not twice (C) twice differentiable but not thrice (D) thrice differentiable 11.

x

0 (A)

(

)

(C)

(

(B)

. /

(D)

. /

)

IN – 2006 7. The function ( ) is approximated as where is in radian. The maximum value of for which the error due to the approximation is within (A) 0.1 rad (C) 0.3 rad (B) 0.2 rad (D) 0.4 rad

()

( )∫

(A) (

()

(

))

(B) (

()

(

))

(

()

(

))

(

()

(

))

()

(

(D) (

(

)

(

) ( )

( )) ( ))

13.

The expression (A) – (B) x

14.

Given y =

(A) √π⁄ (B) √π 10.

(C) Π (D) π⁄

Consider the function f(x) = , where x is real. Then the function f(x) at x = 0 is (A) continuous but not differentiable

+ 2x + 10, the value of

|

(C) 12 (D) 13

15. (A) Indeterminate (B) 0

(C) 1 (D)

IN – 2009 16. A sphere of unit radius is centered at the origin. The unit normal at a point (x, y, z) on the surface of the sphere is (A) (x, y, z) (C) . /

IN – 2007 9. The value of the integral dx dy is.

√

for x > 0 is equal to (C) (D)

is equal to (A) 0 (B) 4

(B) .

∫ ∫

(C) (D)

is.

IN – 2008 12. Consider the function y = x2 6x + 9. The maximum value of y obtained when x varies over the interval 2 to 5 is (A) 1 (C) 4 (B) 3 (D) 9

The solution of the integral equation ()

(C)

For real x, the maximum value of (A) 1 (B) e 1

1

8.

Mathematics

√

√

√

/

(D) .

√

√

√

√

√

√

/

IN – 2010 17. The electric charge density in the region R: is given as σ( ) , where x and y are in meters. The total charge (in coulomb) contained in the region R is (A) π (C) π⁄ (B) π (D) 0 th

th

th


GATE QUESTION BANK

18.

The integral ∫

.

evaluates to (A) 6 (B) 3 19.

/ sin(t) dt

23.

A scalar valued function is defined as ( ) , where A is a symmetric positive definite matrix with dimension n× n ; b and x are vectors of dimension n×1. The minimum value of ( ) will occur when x equals ) (A) ( (C) . / ) (B) – ( (D)

24.

Given ()

(

()

o .

(C) 1.5 (D) 0

The infinite series ( )

…………

converges to (A) cos (x) (B) sin(x)

(C) sinh(x) (D)

IN – 2011 20.

The series ∑ for (A) (B)

(

)

Mathematics

π) π

π /

The o w (A) A circle (B) A multi-loop closed curve (C) Hyperbola (D) An ellipse

converges

(C) (D)

IN – 2013 21. For a vector E, which one the following statement is NOT TRUE? (A) E E o o (B) If E E is called conservative (C) If E E is called irrotational (D) E E -rotational IN – 2014 22. A vector is defined as ̂ ̂ ̂ ̂ are unit vectors in Where ̂ ̂ Cartesian ( ) coordinate system. The surface integral ∯ f.ds over the closed surface S of a cube with vertices having the following coordinates: (0,0,0),(1,0,0),(1,1,0),(0,1,0),(0,0,1), (1,0,1),(1,1,1),(0,1,1) is________

th

th

th


GATE QUESTION BANK

Mathematics


= =

Since, potential function of ⃗ is x²yz ( ) ( ) ( ) 2.

0 o

1

[Ans. A]

[Ans. D] o

h

π h

π o

(

√

0

9.

[Ans. B]

10.

[Ans. B] (

)

For V to be max

)

This is of the form . /

Hence, h

Applying L hospital rule (

3.

√

1

)

[Ans. A]

. /

= (

)

|

|

|

|

= = 11.

4.

[Ans. A] (

After changing order ∫ ∫ 5.

[Ans. A] I= ∫ (

)

=2∫

[ ∫

[Ans. B] Let the vectors be

) ( )(⃗ )(⃗ )

]

= 2∫ 6.

[Ans. A] A Line integral and a surface integral is connected by stokes theorem

7.

[Ans. B]

Now Area vector will be perpendicular to plane of i.e. will be the required unit vector. And option (A), (D) cannot give a vector product )| |(⃗ ⃗ ) (⃗ 12.

[Ans. B]

Applying ’ Hospital rule, we get I= 8.

[Ans. A] ∫

√

Given:

I=

(

)

(

)

√

√

√

For 0 1

[

] th

th

th


GATE QUESTION BANK

13.

14.

But y is always greater than x. Hence y= 4 only.

= ∫

[Ans. B] Since interval given is bounded, so minimum value of functions is 1.

= 0

(

)

)

)

Now by partial fractions, (a3 8) = (a 2)(a2+2a+4)

) |

⇒L=

[Ans. D] To see whether the integrals are bounded or unbounded, we need to see that the o ’ h h interval of integration. Let us write down the range of the integrands in the 4 options, Thus, (D) , i.e., ∫ [Ans. B] h

19.

o

o

o Φ (

Φ)

̂

̂ ). (

̂

( ̂

( )

(

( )

)

( )

( )

Coefficient of (x- )⁴ Now f(x)= ex ⇒ (x)= ex ⇒ (a)= ea ( )

Hence for a=2, 20.

⃗⃗

[Ans. D] div {( (

Hence directional derivative is (grad (x2+2y2+z)).

=

[Ans. C] Taylor series expansion of f(x) about a is given by ( )

dx is unbounded.

along a vector ⃗

(2x ̂

( (

A (0,1); B (0,1); C (0, ); D (0, )

16.

[Ans. B]

Let x= a3 ⇒ a=2

=1.22 15.

1

).dx

= (

)

L=

∫√

L =∫ (√

)

= 18.

h

(

= ∫ (

[Ans. D] h

Mathematics

)̂

(

)

)̂ (

)̂}

( )

(

)

=3

̂)

√ ̂)

21.

[Ans. C]

= Hence at (1,1,2), ⇒

Directional derivative = 17.

[Ans. A] I = ∬ .dx dy The limit of y is form 0 to

and limit

of x from 0 to 2 I =∫ ∫

⇒

( )

⇒

( (

∫

.

)(

) )

/ th

th

th


GATE QUESTION BANK

22.

[Ans. A] Given:

23.

y2  4x x2  4y

Mathematics

[Ans. A] Short method: Take a point on the curve z = 1, x = 0, y=0 Length between origin and this point ) ( ) ( ) =1 √( This is minimum length because all options have length greater than 1.

(4,4) (0,0)

24.

x4   4x 16

[Ans. B] Y

or x4  64x

B

or x(x  64)  0 3

or x3  64 or x  4

x = cos  y=sin 

 y 4  Required area = ∫ .√

Path is x2  y2  1

/

R  e (x  y)2  1  2sin  cos 

4

 2 x3   2 x3 2    3 120  4 64  (4)3 2  3 12 32 16 16    3 3 3

2

2

 cos2    (1  sin2)d   2 0  0 =

Alternately For point where both parabolas cut each other

 1 1     1 2 2 2 2

Alternately Given: x2  y2  1 Put x=cos  , and y=sin 

y2  4x, x2  4y

 x  y 2  cos2  sin2  2sincos

 x  4 4x 2

= 1  sin2

or x2  8 x or x4  64x

∫ (

or x3  64

 x  4,0 ,(4,0)

2

4

x2 dx 0 4

     1 2 

  4x   0

)

 cos2    1 1        2 0  2 2 2  

 Required area 4

X

A

4

 2 x3  16  2 x3 2     3 120 3 

25.

[Ans. C]

F  3xzi  2xyj  yz2k ⃗ ⃗

th

th

th


GATE QUESTION BANK

(

)

(

)

(

√

)

 3z  2x  2yz

π ∫

At point (1, 1, 1), divergence =3+2 2=3 26.

⃗

o

30.

⃗ ̂

̂

||

31. ( ̂

27.

)̂

(

∫ ( )

[Ans. D] Standard limit formulae

) ̂

32.

[Ans. B]

33.

[Ans. C] The function is continuous in [ 1, 1] It is also differentiable in [ 1, 1] except at x = 0. Since Left derivative = 1 and Right derivative = 1 at x = 0

34.

[Ans. B]

[Ans. C]

1

1

2

y is continuous for all x differentiable for all x since at

o

o

R, and R, except at

o

,

Using this standard limit, here a = 1 then = ( ) /2 =1/2

’ h

value towards the left and right side of

35.

[Ans. D] ( ) ( ) ( ) ( ) f(x) has a point of inflection at x =0.

36.

[Ans. A]

[Ans. D] ,

∫

29.

( )

̂

⇒

28.

π

)

[Ans. D] If f(x) even function ∫

||

o

π* +

π(

[Ans. D] ⃗ ̂

Mathematics

-

[Ans. D]

π

∫ π

(

) ̂

Volume from x = 1 to x = 2,

̂

∫π ( √ th

th

√

̂ ̂ ̂

̂

) th


GATE QUESTION BANK

̂

√ √ ̂

√ ̂

̂ ̂

√ ̂

38.

(√ √

37.

̂

̂ √

[Ans. A] By Gauss Divergence theorem, ∬( ̅ ̅)

√ √ The unit outward normal vector at point P is (

Mathematics

)

∭(

(Surface Integral is transformed to volume Integral)

)

( )

( )

( )

)̂

√ ̂

)

∭(

∭

[Ans. A] The area enclosed is shown below as shaded

π π

(

∬( ̅ ̂)

)

)

∭(

( π) (

)

The coordinates of point P and Q is obtained by solving y = x and y = simultaneously, i.e. x = ) ⇒ ( ⇒ Now, x = 0 ⇒ which is point Q(0,0) and x = 1 ⇒ which is point P(1,1) So required area is

π 39.

[Ans. C] ∫(√ ) ( ) Using Integration by parts ∫

∫

Here, f=ln(x) and dg=√ and g=

∫

* +

∫

o ∫(√ ) ( )

* + [

]

∫

[

]

[

(

th

th

( ) ] )

th


GATE QUESTION BANK

40.

π

[Ans. A] o

0

1

So the minimum value is

[Differentiating both o o Hospital method]

o w

.

=

’

. /

45.

o

41.

Mathematics

/

[Ans. D] o o o ( ) ( )

o

( )

o

o

otherwise it is said to be discontinuous. So the most appropriate option is D.

[Ans. B] G

o 46.

|

|

[Ans. C] ̂ ̂ Div

̂

(

)

Vectors are linearly dependent 42.

[Ans. B] (

) ) ( ) , o ( )( ) ( ) o ( )( ) 43.

(

47. -

ho

[Ans. B] Let ∫

(

)

(

( o (

)

[Ans. A] ̂

(

⃗ [ ̂[

(

] )

(

) ∫

̂ ̂

)

48.

)]

)

()

o

̂[ ,̂ ( 44.

(

) (

( ) )̂

∫

|

,

-

)] (

,̂ (

o

[Ans. B] ∫ ∫

̂[

o

∫

)] ̂,

)̂

(

-

∫ (

)

|

)̂ [

[Ans. *] Range 1.00 to 0.94 h o π

,

th

th

] -

,

th

-


GATE QUESTION BANK

CE 1.

a =2a cos i.e, cos [Ans C] G ’ theorem is 

∮(

)

∬(

∮ ((



(

( )

)

= y and

=∫

∫

=∫

,

(

)-

=

×

( )

π . /

|

π . /

π (

π *

[Ans. D] Since the position of rail engine S(t) is continuous and differentiable function according to Lagrange’s mean value theorem more )

(

o ) o

∫(

w

∫

(t) = v(t) =

(

π * (

( )

π

π π

√

| π

√

√

)

√

)

)+

+

)

m/sec kmph

4.

[Ans. C] f = 2 +3

= 126 kmph Where v(t) is the velocity of the rail engine. 3.

∫

∫

(

-

= 2y

=∫

=

∫

)

’ h o I= ∫

)

, ∫

 = xy 

)

(

)

)

⇒

R

Here I = ∮ (

2.

Mathematics

[Ans. D] h ’ o h r=2acos (i) r = a represents a circle with centre ( ) ‘ ’ (ii) r = 2acos represents a circle symmetric about OX with centre at ( ) ‘ ’ The circles are shown in figure below. At h o o o ‘ ’ P y Q π 3

O

A

(

)⃗

= 4xi + 6yj + 2zk At P (2, 1, 3) Directional derivation ̂ ( ) ( ) ( ) √ ( ) ( ) ( ) √ √ 5.

[Ans. A] Potential function,

x

th

th

th


GATE QUESTION BANK

8. Integrating ∫

[Ans. D] Put

∫

wh

Mathematics

o

⇒

√



=√



=√

…… ( )

Now given equation is ……….. (ii) 6.

[Ans. B] Let I(α) =∫ (

h ∫ )

.

/

h

(

dx …( ) =

(

)

(

h

)

√ ) [ from eqn(i)]

=

∫ Then Integrating by parts we get, =

0

=

.

( α

√

h

(

)

o )1

/

√

(

√

(

(

h

)

h

)

= dI = Integrating, I = ( )

α o

h

√

)

h

() ( )

+C=0 C= (α)  ( )

α

π

Now substitute in eqn (ii) we get h h

π

I(0) = But from equation (i), I(0) = ∫ ∫

h

⇒ dx

h

⇒

dx =

h h

Which is the desired form 7.

[Ans. D] ̅=5 +2

√

+ 3y ⃗

(⃗ )

9.

[Ans. C] ̅ ̅=0 ̅ ̅ If ̅ ̅ = 0

= 5y + 4y + 6yz At(1, 1, 1) div ( ) = 5.1 + 4.1 + 6.1.1 = 15

is the correct transformation.

o

 o Since P and Q are non-zero vectors  o 0  th

th

th


GATE QUESTION BANK

10.

[Ans. A] Since the limit is a function of x. We first integrate w.r.t. y and then w.r.t. x )

∫ ∫(

∫

√ √

√

)

[Ans. D] Length of curve f(x) between x = a and x = b is given by ∫√

∫

*

√

√ 13.

∫(

Mathematics

(

)

+ Here,

∫ (

4h … … ( )

= 8h

)

Since ∫ (

and y = h at x =

)

* ( ) *

(As can be seen from equation (i), by substituting x = 0 and x = L/2)

( )+

(Length of cable)

+

√

=∫

.

/ ∫ √

ho 11.

[Ans. B] f = + 3 +2 f = grad f = i

+j

[Ans. A]

15.

[Ans. A] ( )

+k

= i(2x) + j(6y) + k(4z) The gradient at P(1, 2, 1) is = i(2×1) + j(6×2) + k(4 ( )) = 2i + 12j – 4k 12.

14.

[Ans. B] (

)

⃗

⃗ ⃗

̂

⃗

h

Putting,

√ o (

Given,

̂ )

.

/ is the only stationary point.

√ *

+ .

√

th

th

/

th


GATE QUESTION BANK

*

Since the limit is in form of

+ .

*

’ ho and get λ

/

+ .

Since, We have either a maxima or minima at

o ()

⇒λ 18.

/

Also since, r=0 )

o

1

.

/

[Ans. A]

= 8 > 0, the point

(

o

)

,

)

19.

o -

[Ans. A] Area = |̅ ̅ | ̅̅̅̅ ̅̅̅̅ R (

So the optimal value of f(x, y) is a

|

) o

o

(

The minimum value is (

o

π

⇒λ

(

o

, we can use

/

Since,

.

Mathematics

)

(

)

|

minimum equal to 16.

̅̅̅̅ ̅̅̅̅ R ̅̅̅̅ ̅̅̅̅ R

[Ans. B] Let I = ∫

√ √

Since ∫ ( ) I=∫

√ √

…( )

√

√

∫ (

20.

[Ans. B]

21.

[Ans. B]

( )

( )

̅(

)

)

…( )

(i) + (ii)  2I = ∫

√

√

√

√

 2I = ∫

∫

2I = |

o ∫

o

o

 I = a/2 17.

⇒

[Ans. C] For a function f(x) to be continuous, at x=a ( ) ( )

o ∫ (

⇒

)

∫ (

)

If f(x) is continuous at x= π . /

*

λ o

+

[

th

th

]

th


GATE QUESTION BANK

[

]

|

(

Mathematics

)

(

)

(

)

|

Substituting the values we get ( ) ( ) ( ) | | 24. π

π

∫

o

∫

o

∫

o

(

[Ans. 12]

o ) ( )

o

o ( ) ∫

o

( )

[ 22.

⇒ Magnitude of acceleration

]

=√

[Ans. C] ( ⇒

) (

25.

(

)

[Ans. B] We have

) ∫ ( )

⇒ , ow

-

∫ ( )

=1+0=1 Hence correct option is (C) 23.

∫ ( )

∫ ( )

[Ans. A] (4, 3) a (2, 2) b

c

x

( )μ

0.5

(1, 0)

0.5

o ∆ wh o –ordinate points are given is given by

th

th

th


GATE QUESTION BANK

26.

[Ans. A]

=

α Use L – hospital Rule

= 4.

α

=1

[Ans. A] P=∑

o

…

‘ ’

= log x

 1 n 1 r  Cr xr  n 1  x  r0





r 0

r 0

‘ ’



r 0

i 0

5.

= 12

g(i) =i+1

– 24 48

)

+ 37

– 48 x = 0

√

x= =2

[Ans. A] f(x)= |x| Continuity: In other words, f(x) = x o ≥ x for x< 0 Since, = =0 , f (x) is continuous for all real values of x Differentiability:

=

√

96x

48

=

√

√ = 36

Now at x = 0 =

48

0

At 2 ± √ also

0 (using

calculator) There are 3 extrema in this function

( )

)

6.

( )

[Ans. D] Since ∫ ( )

R h So |x| is continuous but not differentiable at x=0 3.

(

x (12 – 48x 48 ) = 0 x = 0 or 12 – 48x – 48 = 0 4x – 4 = 0

∑ ()

(

k

)

[Ans. D] y = 3 – 16

(since r is a dummy variable, r can be replaced by i)

)

k

⇒

  r  1 xr   i  1 xi

(

–1)

)

w h (

  r1 Cr xr   r1 C1xr 

( …

 1 21 r  Cr xr  1  x 2 r0

Putting n=2,

2.

)

= Q=∑

[Ans. B]

(

w h a =1, l=2k 1

P= ( CS 1.

Mathematics

I =∫ =∫



=∫ (

)

 (

) (

)

Since tan (A B) =

[Ans. A] =

⁄ ⁄

th

th

th


GATE QUESTION BANK 

[

]

[

]

Mathematics

∫ 

∫

0

1

0

1

9.

[Ans. B] f(x) = sin x ⇒ ( ) o ( ) ⇒ o π π π [



( (

∫

) )

( (

) )

π ]

( )





∫

∫

At

. /

gives maximum

value =,

)-

(

At

= ln ( sec ) – ln (sec 0) = ln (√ ) = ln (

. /

value

( ) 10.

)–0=

[Ans. A] For x =

7.

[Ans. B] (

8.

)

*

(

(

) [

*(

) +

) + .

/

]

11.

[Ans. C] By Mean value theorem

12.

[Ans. A] Define g(x) = f(x) – f(x + 1) in [0, 1]. g(0) is negative and g(1) is positive. By intermediate value theorem there is €( ) h h g(y) = 0 That is f(y) = f(y + 1) Thus Answer is (A)

13.

[Ans. 2] * w + * w + For min maximum non – common elements must be there ⇒ * + must be common to any 2 elements of V1 ( )minimum value = 2

o o ∫

∫

*

+ [

]

,

-(

o π

, f(x) =

For x = , f(x) = 3 – 1 = 2 For x = 3, f(x) = 2 ( ) ( ) = f(3)

[Ans. D] ∫

gives minimum

π

)

th

th

th


GATE QUESTION BANK

14.

Mathematics

[Ans. 4] ∫ ∫

|

∫∫

π ( ) o π o (π) π Hence option (A) is correct

oπ π o π

∫

∫(

) ECE 1.

[Ans. C]

∫ ∫ (∫

∫ o

)

dy  0 for x< 0 dx dy  0 for x> 0 dx

∫ o

o Substituting the limits π o (π) o ( ) π

2.

[Ans. A] Given,

f x 

∫

 f ' x  

1  e .e  e 1  e 

|

∫∫

3.

= x cos

∫(

x 2

2x





ex

1  ex



2

0

o

)

Let cos = t ⇒ At o π o π o

o

∫ o

x

∫

[Ans. A] ∫

x

[Ans. C]

= π o ( π) π o π = π LHS = I + II = π π π⇒ 15.

ex 1  ex

∫

|

∫

∫ o

∫(

)

∫(

∫

∫

th

th

th

)

|

|

(

)

(

)


GATE QUESTION BANK

8.

Mathematics

[Ans. A]

Given, f  x   x2  x  2

df  x  0 dx 4.

 2x 1=0

[Ans. A] o ’ h o )

∬(

 x

∮

1 2

d2f  x  = 2 ve dx2 So it shows only minima for interval [ 4, 4], it contains a maximum value that will be at x= 4 or x=4  f( 4)=18 and f(+4)=10 

5.

6.

[Ans. D] From vector triple product ( ) ( ) ( ) Here, ( ) ( ) ⇒ ( ) ( ) ( )

[Ans. D]

y  f  x ; x  0, 

[Ans. A] ( )

 f  x0 

For strictly bounded, 0  limy  

2 x  x0  f' x0   x  x0  f'' x0      ......

1

 e  (x  2)(e 2

x0

or 0  lim y  

2

 x  2 )

2

2

2

x 2

So, y  e x is strictly bounded

e ...... 2

  x  22  ......   e2  3  x    2   (Neglecting higher power of x)

7.

9.

10.

lim 0

=

ex  e x ex  ex

x x2 x3 e  1     .......... 1 2 3

11.

x

ex  1 

[Ans. B] Two points on line are ( 1, 0) and (0, 1) Hence line equation is,

 y y  y   2 1 x  c  x2  x1  y  x c y  x 1 … ( )

x x2 x3    .......... 1 2 3

x2 x4   .......... ex  ex 2 4  x x  e e x3 x5 x    .......... 3 5 1

or cot h (x)=

sin  /2 1  sin  /2  lim   0 2    /2 

1  sin  /2 1 = lim  2  0  /2  2

[Ans. C] coth (x)=

[Ans. A]

2 2 5 I   ydx    x  1dx   2.5 2 1 1

1 x

(Since at x=1,y=2)

(Neglecting x2 and higher order)

th

th

th


GATE QUESTION BANK

12.

[Ans. B] Taking f(x, y)= xy, we can show that, xdx+ydy, is exact. So, the value of the integral is independent of path

15.

Mathematics

[Ans. A]

Given : g  x,y   4x3  10y 4 The straight line can be expressed as y=2x Then g(x,y)=4x3+ 10 (2x)4

(0, 1)

 1

1





4 I   4x3  10 2x   dx   4x3  160x4 dx   0 0 1

 4x4 160 5  =   x  33 5 0  4

(1, 0)

)

∫(

∫

[Ans. A] f(x)= + (x)= =0 x=0 (x)= + >0  x R. Hence minimum at x=0 f(0)=1+1=2 Alternatively: For any even function the maxima & minima can be found by A.M. >= GM => exp(x) + exp( x) ≥ 2 Hence minimum value = 2

17.

[Ans. B]

∫

[ |

13.

16.

| ]

[Ans. B] Let f(x)  ex  sinx o ’ 2 x  a  f  x   f a    x  a f'a   f''a  2!

Q

where, a= 2 x    f  x   f     x   f'   f''  2!

Coefficient of (x )2 is

f ''  2

P

f''   ex  sinx |at x   e

 Coefficient of (x )2=0.5 exp () 14.

∫(

)

[Ans. A]

o Thus, ( ( )w o ( )w o ( )w

)w h h h

h

∫

∫

[ |

| ]

o ow ow ow ow th

th

th


GATE QUESTION BANK

18.

21.

[Ans. D] sinx = x = (x – π ) –

y=

(

or

19.

(

)

(

)

sin x = (x – π ) –

or

)

(

)

=1 –

(

) (

= 1

(

) (

)

o

.... ( )

)



...

o

....

(

)

( )



....

=

Therefore, at

22.

∬⃗ ⃗

̂

̂

̂

̂

∯

⃗

∭ ( )∭ and is the position vector)

(

23.

⃗⃗⃗ ⃗⃗⃗

has a maximum.

[Ans. D] Apply the divergence theorem

[Ans. C]

[Ans. A] ̂

Y

S

3

R

1

̅

Q

P

⇒

√

√

⇒

∮ ⃗ ⃗⃗⃗ ∫ ⃗ ⃗⃗⃗

∫ ⃗ ⃗⃗⃗ √

∫ ⃗ ⃗⃗⃗

∫

√

∫ ⃗ ⃗⃗⃗

∫ .√ /

∫ ⃗ ⃗⃗⃗

∫√ √

[ ∫ ⃗ ⃗⃗⃗

√

* +

[

) )

⇒

25.

[Ans. B]

, √ √

( )

)]

( (

)]

[Ans. C] ( ) , ( ) ( ) ( ) ⇒ are the stationary points ( ) ( ) ( ) and f(2) = 25 and f(4)=21 M o ( ) , f(6)=41

√

. /

(

(

24. ]

] ∫ .√ /

[

[

̂

⇒

∫ ⃗ ⃗⃗⃗

along PQ y =1 dy =0]

∫ ⃗ ⃗⃗⃗

⇒

⇒

X

= [

o

Since

[Ans. D] o ’ h o ⃗ ⃗ = ∮

⃗⃗⃗

 o

o

According Stokes Theorem ⃗⃗⃗ ⃗ ∮ ⃗ ⃗ =∮

20.

[Ans. A]

....

sin (x –π )

or

Mathematics

∮ ⃗ ⃗⃗⃗ ⇒ th

th

th


GATE QUESTION BANK

o

30.

[Ans. C]

E o

(

E o

31.

o

[Ans. *] Range ( ) ( )

̂

̂

⇒ ⇒ ( )

=1+1+1 =3 [Ans. D] o ’ h o “ h integral of a ⃗ vector around a closed path L is equal to the integral of curl of ⃗ over the open ∮⃗

h

⃗

∬(

o

∫

∫

∫

∫

(

∫

*

32.

h ”

33.

)

) ) ) )

[Ans. C] Let x (opposite side), y (adjacent side) and z (hypotenuse side) of a right angled triangle

+

∫

29.

[Ans. *] Range 5.9 to 6.1 Maximum value is 6 ( ) ( ) ( ( ( (

⃗ )⃗

[Ans. *] Range 862 to 866 Volume under the surface ∫

( ) ( ) h

o

o

28.

to 0.01

( )

[Ans. D] ̅ ̂

=

27.

)

π

⇒

26.

Mathematics

Given

o

… )(

(

[Ans. A] o ( ) ̇( )

o

⇒ o

⇒ ( ) Since ( ) is negative, maximum value of f(x) will be where ( )

⇒

o 0(

⇒ ⇒

)

o ⇒ ( )

⇒

o

( )

( )

o

oh

(

)

)

1

( ( (

th

th

)(

))

) th


GATE QUESTION BANK

By trial and error method using options π

34.

Now at x = 2 (2) = ( ) = ( ) = 2 <0 At x = 2 we have a maxima.

[Ans. *] Range 6.8 to 7.2 ⃗ ̂ ̂ ̂ ̂

̂

=

̂

̂

̂

3.

=∫

[Ans. *] Range 2.99 to 3.01 √

⃗

√

)

√

(

(

)

)̂

At (1, 1), ⃗

√

(̂

Given unit vector, ̂

(

√

√

)̂

(̂

√

√

.

-

,

-

/

[Ans. D]

5.

[Ans. D] We consider options (A) and (D) only because which contains variable r. By integrating (D), we get π , which is volume of cone.

6.

[Ans. D] By property of definite integral

̂ ) ̂ )

=3

) ∫ ( ) ∫ ( π On simplification we get option (D)

[Ans. C] Grad u = ̂

⁄

At (1, 3) Grad u = √

,( ⁄ )

7.

[Ans. B] f(x) = ( ) (x) = 2( ) =4x( ) =0 x = 0, x = 2 and x = 2 are the stationary points. (x) = 4[x(2x) +( ) ] = 4[2 = 4 [3 = 12 (0) = < 0, maxima at x = 0 (2) =(12) = 32 > 0, minima at x = ( 2) =12( ) = 32 > 0; minima at x = There is only one maxima and only two minima for this function.

-

=√ 2.

,

4.

̂ )

So, directional derivative ⇒ ⃗ ̂ (̂ ̂ ) (̂

EE 1.

1

[Ans. C]

=

(

0

̂

At (1, 1, 1) ⃗ |⃗ | √

35.

Mathematics

[Ans. A] f(x) = (x) = ( ) = ( ) Putting ( (x) = 0 ( )=0 ( )=0 x = 0 or x = 2 are the stationary points. Now, ( ) ( (x) = )( ) = ( ( )) = ( ) ( )=2 At x = 0, (0) = Since (x) = 2 is > 0 at x = 0 we have a minima. th

th

th


GATE QUESTION BANK

8.

Mathematics

= , =, ( = (

[Ans. A]

))

(

) =1

1

14.

[Ans. B] Dot product of two vectors =1+a+ =0 So orthogonal

15.

[Ans. C] f(x) = ( ) ( ) So the equation f(x) having only maxima at x = 1

16.

[Ans. B]

0

√

∫

9.

10.

(

∫

)

[Ans. C] ( ) ( ) ( ) ( ) ( )

̂

[Ans. D] ̅=( )̂ ( ( ) = (0, 2) ( ) = (2, 0) Equation of starting line

̂ ̂ ∫

11.

) y = 2 – x and dy = – dx

17.

C

)

o ( )

o

.(

)̂

̂

(

)̂

̂

̂ ||

( ̂ ( =0

⃗ ̂

̂

||

̂ ̂

)̂ /

̂

is undefined

[Ans. A] ̂ Div ( ) =.

‘ ’

(

Discontinuous

/(

̂

̂

̂)

18.

= 1+1+1= 3 13.

[Ans. D]

)

But at

12.

∫

[Ans. B] (

̂

Along x axis ,y=0,z=0 The integral reduces to zero.

=

∫ (

̂ ̂

̂

)̂

⇒ y = 2 – x , dy = – dx ̅ ̅ =( ( ) Putting ∫̅ ̅ ∫

̇̂

[Ans. B] P=∫ th

[Ans. A] ( ) o ⇒ M th

) ( ) ̂ (

( ( )

th

) ( ) ̂ (

)

(

) (

)


) )

GATE QUESTION BANK

19.

Mathematics

[Ans. B]

.

0

π

/1

∫⃗ ∫

[

(

o

o π

)( )(

∫ ( ∫

⇒

o

23.

[Ans. A] ( ⃗) ⃗ ⃗ ( ) ( ) ⃗

24.

[Ans. B] ( ) ( ) ⇒ ⇒ )( ⇒( ⇒ ( ) ( ) ( )

) )

[

]

π 20.

]

[Ans. C] ( )

(

)

( )

( ) For number of values of ) o ( ) ( ( ) ( ( )

( ) ⁄

,

w

-

) ( ) ( )

M 21.

)

[Ans. B]

IN 1.

G o

h

[Ans. A]

o

o

⇒ o

⇒

(

o

)

Unit vector along y = x is G

∫ (∫ ∫ (∫

22.

[Ans. 2] ( ∫

)

̂

)

̂

π

√ o

√ ̂

.

) ∫

π

o

o .

π

/

.

π

/.

√

/

√

/

√

√

√ √

2.

[Ans. D] Using L Hospital Rule., numerator becomes =

From the graph, distance at

th

th

()

= ( )

th


GATE QUESTION BANK

3.

[Ans. B]

Mathematics

When

Given integral is, I=∫

( )

Let f(x) = so curve of 1/

will be

(

)

(

)

And when

f(x)

( )

The possible expression for f(x) is 1

. 7.

-1

0

1

/

[Ans. B]

x

Error,

This curve will be discontinuous at x=0 o ’ w o

For error to be minimum (

4.

[Ans. A] ̅ (t) =x (t) ̂+y (t) + Let R ̂ z (t) ̂ ̅( ) =K (constant) |R i.e., (t) + (t) + (t) = constant. On analyzing the given (A) option, we find ̅(t) that R

̅( )

⇒ ⇒

[Ans. C] Given : f= + where,

(

)

⇒

√

will give constant magnitude, √

1

G …… + (i=0 to n) are constant.

=

+(n 1)

o

…… ⇒

+ and

)

o

so first differentiation of the integration will be zero. 5.

o

=0+

+

(n 1)

√

…… ⇒

+n

√

+ = , = nf 6.

+

+

⇒

-

(

)

⇒ ⇒

[Ans. B] ( )

(

)

8.

[Ans. B] ()

When ( )

(

)

(

)

( )

∫

…( )

Differentiating the above equation

When ( )

th

th

th


GATE QUESTION BANK

()

,

14.

() ∫

Mathematics

[Ans. B] Given y = x2 + 2x + 10 = 2x + 2

( ) -

| From equation (i) () ()

⇒

()

()

This is Leibnitz linear equation Integrating factor I.F = ∫ the solution is ()

15.

[Ans. C] By definition

16.

[Ans. A]

()

Unit vector=

=xi+yj+zk

and 17.

∫

[Ans. C] R: Y

( ) 1 1

, [Ans. D]

10.

[Ans. A] This is a standard question of differentiability & continuity

Area =

[Ans. C] y= =(

X

- o

9.

11.

+1

( )

Total charge = σ = = 18.

).(cos x + sin x) = 0

⇒ tan x = 1 Or x =

coulomb.

[Ans. B] We know that ∫

() (

∫

.

) π

( )wh π . /

/

y will be maximum at x = y=

19.

= 12.

13.

[Ans. C] y(2) = y(5) =

=

√

[Ans. B] Expansion of sin x ........

( ) ( )

20.

[Ans. B] In a G.P

(

)

For a G.P to converge

[Ans. C] y= y=

(

⇒

)

⇒

(

)

⇒ th

th

th


GATE QUESTION BANK

21.

[Ans. D] .E=0 is not irrational (it is solenoidal)

22.

[Ans. 1] From Gauss divergence theorem, we have ∫ ̅ ̅

̅

∫

/dxdydz

∫ [ ⇒

̅

∫

∫.

Mathematics

∫ ̅

[Ans. C]

24.

[Ans. D]

̂

) ̂

̂

]

o .

π

π /

23.

(

∫

(

π )

th

th

th


GATE QUESTION BANK

Mathematics

Differential Equations ME – 2005 1.

If x

xy

n y

what is y (A) e (B) 1

then

⁄ ⁄

(C) (D)

(B) degree 1 order 1 (C) degree 2 order 1 (D) degree 2 order 2 ME – 2007 7.

2.

3.

Statement for Linked Answer Questions 2 and 3. The complete solution of the ordinary differential equation y y p qy s x x y Then, p and q are (A) p =3, q = 3 (C) p =4, q = 3 (B) p =3, q = 4 (D) p =4, q = 4 Which of the following is a solution of the differential equation y y p q y x x (A) (C) x (B) x (D) x

The solution of

For

ME – 2008 8. It is given that + 2y + y = 0, y (0) = 0, y(1) = 0. What is y (0.5)? (A) 0 (C) 0.62 (B) 0.37 (D) 1.13 9.

Given that ẍ + 3x = 0, and x(0) = 1, ẋ (0) = 0, what is x(1)? (A) 0.99 (C) 0.16 (B) 0.16 (D) 0.99

ME – 2009 10.

+ 3y =

, the particular

integral is: (A) (B) (C) (D) 5.

The solution of x

y

x

with the

s

(A) y

(C) y

(B) y

(D) y

ME – 2010 11. +

The solution of the differential equation

(A) (1+ x)

(C) (1 x)

(B) (1+ x)

(D) (1

The Blasius equation,

, is a

(A) second order non-linear ordinary differential equation (B) third order non-linear ordinary differential equation (C) third order linear ordinary differential equation (D) mixed order non-linear ordinary differential equation

2 dy  2xy  ex with y (0) = 1 is: dx

6.

(D) 2  x  2

(B)   x  1

condition y +4

with initial value

y (0) = 1 is bounded in the interval (C) x  1,x  1 (A)  x 

ME – 2006 4.

y

x)

The partial differential equation (

)

(

)= 0 has

(A) degree 1 order 2 th

th

th


GATE QUESTION BANK

ME – 2011 12. Consider

the

differential

x ,yt x ,yt

equation

y x. The general solution with constant c is (A) y

t n

(B) y

t n (

(C) y

t n ( )

(D) y

t n(

17.

)

18.

differential y

equation

with the boundary

(B) s n ( )

(D)

partial u

15.

19.

s n( )

differential

equation

is a

linear equation of order 2 non – linear equation of order 1 linear equation of order 1 non – linear equation of order 2

subjected to the boundary conditions u(0) = 0 and u(L) = U, is (A) u (C) u ( ) (

)

(D) u

(

x

y and

x

x ,y-

*

x + ,y-

*

x + ,y-

y is

20.

with t __________

The general solution of the differential os x

constant, is (A) y s n x

y

(B) t n (

)

y

(C)

os (

)

x

(D) t n (

)

x

y with c as a x

Consider two solution x(t) = x t and x t x t of the differential equation x t x t t su t t t x t x | t x t | t

t s (A) 1 (B) 1

x t

x t

| t

(C) 0 (D)

The solution of the initial value problem xy y

)

ME – 2014 16. The matrix form of the linear system

x ,yt

at x

The wronskian W(t) =|

where k is a constant,

t

x + ,y-

If y = f(x) is the solution of

x

The solution to the differential equation

(B) u

*

x

conditions of y(0) =0 and y(1) = 1. The complete solution of the differential equation is (A) x (C) s n( )

(A) (B) (C) (D)

x + ,y-

the boundary conditions y

)

the

x

ME – 2013 14. The

*

equation

ME – 2012 13. Consider x

t n

Mathematics

is

(A)

(C)

(B)

(D)

CE – 2005 1. Transformation to substituting v = y

linear form by of the equation

+ p(t)y = q(t)y ; n > 0 will be (A) th

+ (1 n)pv = (1 n)q th

th


GATE QUESTION BANK

2.

(B)

+ (1 n)pv = (1+n)q

(C)

+ (1+n)pv = (1 n)q

(D)

+ (1+n)pv = (1+n)q

CE – 2007 6. The degree of the differential equation + 2x = 0 is (A) 0 (B) 1

in the range (A) (B)

y x

( os x ( os x

(C)

( os x

(D)

( os x

,

( )

7.

The solution for the differential equation = x y with the condition that y = 1 at

is given by

x = 0 is

s n x)

(B) In(y) =

s n x) s n x)

8.

xy

x

+4

(D) y =

A body originally at 600C cools down to C in 15 minutes when kept in air at a temperature of 250C. What will be the temperature of the body at the end of 30 minutes? (A) 35.20C (C) 28.70C 0 (B) 31.5 C (D) 150C

CE – 2008 9.

The general solution of (A) (B) (C) (D)

The solution of the differential equation x

(C) In(y) =

(A) y =

s n x)

CE – 2006 3. A spherical naphthalene ball expanded to the atmosphere losses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in (A) 6 months (C) 12 months (B) 9 months (D) Infinite time

+ y = 0 is

y = P cos x + Q sin x y = P cos x y = P sin x y=Psn x

given that at x = 1, 10.

y = 0 is

5.

(C) 2 (D) 3

The solution of y

4.

Mathematics

(A)

(C)

(B)

(D)


= 0.25 y is to be

solv us ng t b w r mpl t Eul r’s method with the boundary condition y = 1 at x = 0 and with a step size of 1. What would be the value of y at x = 1? (A) 1.33 (C) 2.00 (B) 1.67 (D) 2.33

Solution of (A) x (B) x

=

at x = 1 and y = √ is

y y

(C) x (D) x

y y

CE – 2009 11. Solution of the differential equation 3y

+ 2x = 0 represents a family of

(A) Ellipses (B) Parabolas

(C) circles (D) hyperbolas

CE – 2010 12. The order and degree of the differential equation

+ 4 √( )

respectively (A) 3 and 2 (B) 2 and 3 th

th

y

= 0 are

(C) 3 and 3 (D) 3 and 1 th


GATE QUESTION BANK

13.

The solution to the ordinary differential equation (A) (B) (C) (D)

14.

+

y= y= y= y=

The following differential equation has 3

d2y  dy   4   y2  2  x dt2  dt  (A) degree=2, order=1 (B) degree=1, order=2 (C) degree=4, order=3 (D) degree=2, order=3

6y = 0 is

3

+ + + +

The partial differential equation that can be formed from z = ax + by + ab has the form (w t p (A) (B) (C) (D)

2.

n q

Mathematics

)

ECE – 2006

3.

For the differential equation

z = px + qy z = px + pq z = px + qy + pq z = qx + pq

the boundary conditions are (i) y=0 for x=0 and (ii) y=0 for x=a The form of non-zero solutions of y (where m varies over all integers) are m x y ∑ sn

CE – 2011 15. The solution of the differential equation + = x, with the condition that y = 1 at x = 1, is (A) y =

+

(D) y =

+

CE – 2012 16. The solution of the ordinary differential y=0 for the boundary

condition, y=5 at x = 1 is (A) y (C) y (B) y (D) y CE – 2014 17. The

y

∑

y

∑

y

∑

os

m x

(C) y = +

(B) y = +

equation

d2y  k2y  0 2 dx

integrating

for

the

equation (A) (B)

x

ECE – 2007 4. The solution of the differential equation

d2y  y  y 2 under the boundary dx2 conditions (i) y=y1 at x=0 and (ii) y=y2 at x=, where k, y1 and y2 are constants, is (A) y y y xp( x⁄ ) y (B) y y y xp x⁄ y (C) y y y s n x⁄ y (D) y y y xp x⁄ y k2

differential s

(C) (D)

ECE – 2005 1. A solution of the following differential equation is given by

d2y dy  5  6y  0 dx dx2

ECE – 2008 5. Which of the following is a solution to the differential equation

2x 3x (A) y  e  e

2x 3x (C) y  e  e

2x 3x (B) y  e  e

2x 3x (D) y  e  e

(A) t (B) x t

th

th

x t

x t

(C) x t (D) x t

th

t t


GATE QUESTION BANK

ECE – 2009 6. The order of the differential equation

ECE– 2011 10. The solution of the differential equation

3

d2y  dy    y4  et is dt 2  dt  (A) 1 (C) 3 (B) 2 (D) 4

7.

Match each differential equation in Group I to its family of solution curves from Group II. Group I Group II 1. Circles dy y  P.

dx x dy y  Q. dx x

(A) (B) (C) (D)

3. Hyperbolas

dy x  dx y

y x

(C) y (D) y

ECE\EE\IN – 2012 11. With initial condition x(1) = 0.5, the solution of the differential equation, t

x

t is

(A) x

t

(C) xt

(B) x

t

(D) x

ECE\IN – 2012 12. Consider the differential equation y t y t y t t t t wt y t | n | num r

l v lu o

(A) (B)

x with the initial condition

y s ng Eul r’s rst or r m t o with a step size of 0.1, the value of y is (A) 0.01 (C) 0.0631 (B) 0.031 (D) 0.1 A function n x satisfies the differential equation

is

(A) x (B) x

P-2, Q-3, R-3, S-1 P-1, Q-3, R-2, S-1 P-2, Q-1, R-3, S-3 P-3, Q-2, R-1, S-2

ECE – 2010 8. Consider a differential equation

9.

y y

2. Straight Lines

dy x R.  dx y S.

Mathematics

where L is a

constant. The boundary conditions are: n and n . The solution to this equation is (A) n x xp x (B) n x xp x √ (C) n x xp x (D) n x xp x

y | t (C) (D) 1

s

ECE – 2013 13. A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y t or t when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes – 2y(t) for t > 0, we need to (A) Change the initial condition to – y(0) and the forcing function to 2x(t) (B) Change the initial condition to 2y(0) and the forcing function to –x(t) (C) Change the initial condition to j√ y(0) and the forcing function to j√ x(t) (D) Change the initial condition to – 2y (0) and the forcing function to – 2x(t)

th

th

th


GATE QUESTION BANK

ECE – 2014 14. If the characteristic equation of the differential equation y t

15.

has two equal roots,

n t v lu s o (A) ±1 (B) 0,0

r (C) ±j (D) ±1/2

xy

(C)

(B)

xy

(D)

(B) x t (C) x t (D) x t EE – 2011 3. With K as a constant, the possible solution for the first order differential equation is

Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively? (A)

Mathematics

xy

(C) (D)

(A) (B)

EE – 2013 4. A function y x x is defined over an open interval x = (1,2). At least at one point in this interval ,

16.

17.

18.

If z

xy ln xy then

(A) x

y

(C) x

y

(B) y

x

(D) y

x

If a and b are constants, the most general solution of the differential equation x x x s t t (A) (C) bt (B) bt (D)

(A) 20 (B) 25

tx

s

6.

x x (A) x t

n

with initial conditions | t

, the solution is

(B) s n t

os t

(C) s n t

os t

(D) os t

t

Consider

the

x

EE – 2005 1. The solution of the first order differential qu t on x’ t 3x(t), x (0) = x is (A) x (t) = x (C) x (t) = x (B) x (t) = x (D) x (t) = x EE – 2010 2. For the differential equation

(C) 30 (D) 35

EE – 2014 5. The solution for the differential equation x x w t n t l on t ons x t x n | s t (A) t t

With initial values y(0) = y (0) = 1, the solution of the differential equation y

is exactly

x

differential

equation

y

Which of the following is a solution to this differential equation for x > 0? (A) (C) x (D) ln x (B) x IN– 2005 1. The general solution of the differential equation (D2 4D +4)y = 0, is of the form (given D = d/dx), and C1 and C2 are constants (A) C1e2x (C) C1e2x + C2 e2x 2x (B) C1e + C2 (D) C1e2x + C2x th

th

th


GATE QUESTION BANK

2.

urv s or w t urv tur ρ t 3 any point is equal to cos θ w r θ s t angle made by the tangent at that point with the positive direction of the x-axis, r

gv nρ

⁄

, where y and y

are the first and second derivatives of y with respect to x) (A) circles (C) ellipses (B) parabolas (D) hyperbolas IN– 2006 3. For an initial value problem ÿ ẏ y y n ẏ various solutions are written in the following groups. Match the type of solution with the correct expression. Group 1 Group 2 P. General solution 1. 0.1ex of homogeneous equations Q. Particular integral 2. (A cos 10 x + B sin 10 x) R. Total solution 3. cos 10 x + x satisfying boundary 0.1e conditions (A) P-2, Q-1, R-3 (C) P-1, Q-2, R-3 (B) P-1, Q-3, R-2 (D) P-3, Q-2, R-1 4.

A linear ordinary differential equation is given as

d2y dy  3  2y  δ(t) 2 dt dt Where  (t) is an impulse input. The solut on s oun by Eul r’s orw r difference method that uses an integration step h. What is a suitable value of h? (A) 2.0 (C) 1.0 (B) 1.5 (D) 0.2

Mathematics

IN– 2007 5. The boundary-value problem y λy y y w ll v non-zero solut on n only t v lu s o λ r (A) ± ± … (B) … (C) … (D) … IN– 2008 6. Consider the differential equation = 1 + y2. Which one of the following can be a particular solution of this differential equation? (A) y = tan (x + 3) (C) x = tan (y + 3) (B) y = tan x + 3 (D) x = tan y + 3 IN– 2010 7. Consider y

the

differential

equation

with y(0)=1. The value of

y(1) is (A)

(C)

(B)

(D)

IN – 2011 8. Consider the differential equation ÿ ẏ y with boundary conditions y(0) = 1, y(1) = 0. The value of y(2) is (A) 1 (C) – (B) (D) IN– 2013 9. The type of the partial differential equation

is

(A) Parabolic (B) Elliptic 10.

th

(C) Hyperbolic (D) Nonlinear

The maximum value of the solution y(t) of the differential equation y t ÿ t with initial conditions ẏ and y , for t is (A) 1 (C) (B) 2 (D) √

th

th


GATE QUESTION BANK

Mathematics

IN– 2014 11. The figure shows the plot of y as a function of x

y

y

x

x

The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is : (A) (B)

x

(C)

x

(D)

|x|

th

th

th


GATE QUESTION BANK

Mathematics


4. [Ans. D] x y y

lnx x

xy x

lnx x

y

omp r ng w g t ow y(I.F.) = ∫ x y

∫

olv ng bov utt ng x x 2.

3.

[Ans. B] The given differential equation may be written as y y y ux l ry qu t on s

w

lnx x

x ∫

∫

x

Substituting D=2, we get

x x

(

x

n t v lu o t n n t v lu o y t

5.

[Ans. B] First order equation,

sy

y

dy  Py  Q, dx

Where P = 2x and Q = Since P and Q are functions of x, then Integrating factor,

[Ans. C] Given equation is y p qy x x p q y p q ts solut on s y um o roots p p ro u t o roots q q [Ans. C] Given equation is y y p q x x p q ut p n q y

)

2

I.F. = e  Pdx  e x Solution is y

∫

y

x

∫

x

2

yex  x  c ,c=1

Since, y

 x2

(1 + x) e

y 6.

[Ans. A] Order: The order of a differential equation is the order of the highest derivative appears in the equation Degree: The degree of a differential equation is the degree of the highest order differential coefficient or derivative, when the differential coefficients are free from radicals and fraction. The general solution of differential qu t on o or r ‘n’ must nvolv ‘n’ arbitrary constant.

y

x

th

th

th


GATE QUESTION BANK

7.

y ( )y x … x x Standard form y y … x Where P and Q function of x only and solution is given by

[Ans. C] y x

v n y

y x

y

nt gr t ng y

y

x

∫

x nx

y

n

x

x

x

olut on y x x

and x

Given condition y m ns t x

y

₂ r or yx

y [Ans. D] ẍ x Auxiliary equation is m2 + 3 = 0 i.e. m = ±√ x os√ t sn√ t ẋ os√ t s n√ t √ At t = 0 1=A 0=B x = os √ t x

11.

x

[Ans. B] is third order ( is linear, since the product

) and it is not

allowed in linear differential equation 12.

os √ t

[Ans. D] y x y ∫ y t n

10.

x

x

y 9.

∫x x x

x

yx [Ans. A] y y y A.E is, D2+2D+1 =0 2=0 m 1 The C.F. is (C1+C2x)e-x P.I. = 0 ow y ₁ n y ₂

∫

x

∫

x x

8.

x

Where, integrating factor (I.F) r

y

Mathematics

[Ans. A] Given differential equation is y x y x x

y

th

y t n.

th

y x ∫ x x x

/

th


GATE QUESTION BANK

13.

[Ans. A] 17. y y x x y x x y n y Choice (A) satisfies the initial condition as well as equation as shown below y x y n y y lso x x y x 18. y y x x y x x x x x x x x x o y x is the solution to this equation with given boundary conditions

14.

[Ans. D]

15.

[Ans. B] m m u u At x=0, At x=L, (

[Ans. *] Range 34 to 36 y x y x y x tx y y tx x y x tx y [Ans. D] y os x y x Let x y z y z x x z os z x z os z x z s ( ) z x

z os ( )

Integrating z t n( ) x z t n( ) x x t n(

) n

19.

u x

Solving we get u = U( 16.

Mathematics

)

[Ans. A] x x y t y x y t So by observation it is understood that, x x ,y- * + ,yt

y )

[Ans. A] Since the determinant of wronskian matrix is constant values for, therefore it is same for both t=0 and t= t

20.

x

x t

x t t

[Ans. B] y ∫ ∫ x x y y ln ( ) x y

x t t

x t

ln y

x

ln

v ny n th

y th

th


GATE QUESTION BANK

CE 1.

y tx x

[Ans. A] Given

+ p(t)y = q(t) y

n y

3.

Multiplying by (1 n) y we get v p t n y q t t Now since y = v we get v n pv n q t Where p is p (t) and q is q(t)

t Where, V =

n

4 r

=

sn x os x sn x os x sn x

os x

r )

 r

t utt ng r

n

y

r

r t  dr = kdt Integration we get r = kt + C At t = 0, r = 1 1= k×0+C C=1 r = kt + 1 Now at t = 3 months r = 0.5 cm 0.5 = k × 3 + 1

)

t

…

r r t t Substituting in (i) we get

±

os x

sn x)

r

y (

( os x

[Ans. A]

[Ans. A] y y y x x y y ( ) x This is a linear differential equation

n

s

+ p(t) y = q(t) y ; n > 0

Given, v = y v y n y t t y v t n y t Substitution in the differential equation we get

2.

Mathematics

n solv ng g v s t

sn x

 t = 6 months

y os

sn

4. sn x os x

[Ans. A] Given y x xy – x x y x xy x x Dividing by x

os x

sn x sn x os x

th

th

x

th

y


GATE QUESTION BANK

y x ( )y ( ) x x x Which is a linear first order differential of the form y y x Integrating factor = I.F = ∫ = ∫ y × I.F = ∫ .(I.F)dx x yx ∫( )x x x Now at x = 1, y = 0

Hence, here the degree is 1, which is power of

7.

[Ans. D] y x y x This is variable separable form

x

= x dx

∫ x

∫

x

y y

∫x

tx log y

C x y

5.

–x x

[Ans. B] =

x

y

0C

= Now at t = 30 minutes Θ

±√

y =2

6.

t

 ln θ θ = kt + θ θ C. θ θ C. Given θ = 250C Now t t θ 60 = 25 + C.e0 C = 35 θ At t m nut s θ 40 = 25 + 35

y

y +1=0

t =∫

∫

y y y  0.25hy y +y =0 Putting k = 0 in above equation 0.25h y y +y =0 Since, y = 1 and h = 1 0.25 y

θ θ0) (Newton’s law of cooling)

θ θ θ

[Ans. C] y y y tx x h=1 Iterative equation for backward (implicit) Euler methods for above equation would be y

y x

y

8. y

x

x

log y

i.e. 0 ×

x

Mathematics

= 25 + 35 (

)

= 25 + 35 × ( ) (s n

[Ans. B] Degree of a differential equation is the power of its highest order derivative after the differential equation is made free of radicals and fractions if any, in derivative power.

)

= 31. ≈ C

th

th

th


GATE QUESTION BANK

9.

12.

[Ans. A] +y=0

10.

y

.

13.

∫ x x

√ C=2 Solution is y x

[Ans. D] y y x n y x x This is a linear differential equation of the form y y wt n x x x IF = Integrations factor

x x ∫

y x

y

( )

( )

x

x (

x x

x

y

(

∫

∫

x

Solution is y (IF) = ∫ x  y. x = ∫ xx x  yx = ∫ x x

y )

+

15.

x x y

[Ans. C] y y x x Auxiliary equation is +D–6=0 (D 2) = 0 D = 3 or D = 2 Solution is y =

[Ans. C] Z = ax + by + ab … z p x z q b y Substituting a and b in (i) in terms of p and q we get z = px + qy + pq

[Ans. A]

∫ y y

y y / 0( ) y 1 x x The order is 3 since highest differential

14.

x +y =4

y x y y

y

is

x

3y

y ) x

Removing radicals we get

At x = 1, y = √

11.

√(

The degree is 2 since power of highest differential is 2

[Ans. D] y x x y y dy = x dx ∫y y

[Ans. A] y x

+1=0 E sm m ± General solution is y= [ cos (1 × x) + sin (1 × x)] = cosx + sinx = P cosx + Q sinx Where P and Q are some constants

Mathematics

)

Which is the equation of a family of ellipses

 yx =

th

th

+C

th


GATE QUESTION BANK

y=

 sin ka=0 m x 

+

Now y(1) = 1  ot

 solut on s y

y

17.

m x

[Ans. D] k2D2y= y

y2

y2  2 1  D  k2  y  k2   1 m1 =  k x/k x/k C.F. = C1e  C2 e

[Ans. D]

x/k x/k  y2 y= C1e  C2 e

∫

At y=y1, x=0  y1 = C1+ C2+y2 … At y=y2 , x=  Hence C1 must be zero  y1 = C2+y2  C2 =y1 - y2

[Ans. B]

d2y dy  5  6y  0 dx dx2 A.E. is D2 5D  6  0  D=2,3 2x 3x Hence, solution is y  e  e

2.

sn

x

[Ans. D] y y x y y y

Particular integral (P.I) = = ECE 1.

∑

x 4.

16.

Mathematics

x  y=(y1 – y2) exp    + y2  k 5.

[Ans. B] x t x t t (D +3) x(t) = 0

[Ans. B] 3

d2y  dy   4   y2  2  x dt2  dt  Order of highest derivative=2 Hence, most appropriate answer is (B) 3

3.

[Ans. A] Given, Differential equation,

d2y  k2y  0 dx2 Auxilary equation is y ± Let y os x sn x At x=0, y = 0  A=0 y sn x At x=a, y=0  B sin ka=0 B0 otherwise y=0 always

So, x  t   ke3t , Hence x  t   2e3t is one solution (for some boundary / initial condition) 6.

[Ans. B] The order of a differential equation is the order of the highest derivative involving in equation, so answer is 2. The degree of a differential equation is the degree of the highest derivative involving in equation, so answer is 1.

7.

[Ans. A] P.

∫

∫

log y log x log y xw s qu t on o str g t l n th

th

th


GATE QUESTION BANK

Q.

∫ log y y

x

log x

∫y y y

x

8.

yp rbol

y

∫x x

9.

t ∫t t

xt

t

xt 12.

x

y

[Ans. D] Approach 1: y t y t y t t t t Converting to s-domain s y s sy y sy s y s s s y s s s y s s s n nv rs pl tr ns orm y t t u t y t t t y t | t

y

[Ans. D] Approach 2: y t y t y t t t t Applying Laplace Transform on both sides y s y s sy | t (sy s y ) y s s y s s sy s y s s s y s s s s s

n x m

Auxiliary equation m olut on n x Since, n

±



Since, n  must be zero) Therefore

(hence

The solution is, n x 10.

∫

r l

y old y +0.1 ( ) new x y x y 0 0 0+0 0+0.1×0=0 =0 0. 0 0.1+0 0+0.1×0.1=0.01 1 =0.1 0. 0.0 0.2+0.01 0.01+0.21×0.1 2 =0.21 1 =0.031 The value of y at x= 0.3 is 0.031. x

x=1

Using initial condition, at t = 1, x = 0.5

ypr bol

… Equ t on o

[Ans. B] y x y x x  x y

t

ol s xt

qu t on o

∫y y x

x

∫x x

…

S.

[Ans. C] t

log

… qu t on o

R.

11.

∫

Mathematics

[Ans. C] Given y ln y When y y

and x

y

y t

x

t

t

t

y

y t y t t

y

t

t t t t

th

th

th


GATE QUESTION BANK

z y

x ln xy

ox

z x

y t | t 13.

[Ans. D] Let the differential equation be y t y t x t t Apply Laplace transform on both sides y t {x t } 2 y t 3 t sy s y y s x s s y s x s y x s y y s s s Taking inverse Laplace on both sides x s {y s } 2 3 y { } s s y t x t y So if we want y t as a solution both x(t) and y(0) has to be multiplied by . Hence change x(t) by x t and y(0) by y

14.

[Ans. A] y y y x x The auxiliary equation is m m ± then either m or m i.e., roots of the equation are equal to or

y

z y

18.

EE 1.

xy ln xy

xy

[Ans. C] z xy ln xy z y ln xy x

y

y ln xy

…

x (t) x

x ∫ t x lnx = t x Putting x Now putting initial condition x(0) = x x x Solution is x = x i.e. x(t) = x

0 is a first order linear

xy

.

∫

omog n ous

xy

xy

t

equation (homogeneous) r non l n r qu t ons 16.

x

xy

[Ans. *] Range 0.53 to 0.55 E m m m olut ons s y bx y bx b … s ng y y n gv s n b y x tx y

[Ans. A] v n x’ t

is a first order linear

equation non

x ln xy

[Ans. B] x x x t t Pre auxiliary equation is m m Pre roots of AE are m Repeated roots are present. So, most general solution in n t bt

[Ans. A] xy

x

xy

z y

y

i.e. 15.

xy

xy ln xy

z x

ox 17.

Mathematics

y

th

th

th


GATE QUESTION BANK

2.

[Ans. B] x x x t t Auxiliary equation m m m m (m+4)(m+2)=0 m= 2, 4 x(t) =

6.

[Ans. C] x y xy y y x

m

t

nx

 … On solving (1) & (2), we get

IN 1.

y

x

x( s ts

x

)

x

s

[Ans. C] y

and

m

x(t)= 2 3.

x

x

… (1)



|

y y x

w subst tut y

n x(0) = 1 1=

Mathematics

Since there is double root at 2, so general solution of the given differential equation would be +

[Ans. A] y

m

x

Integrate on both sides 2.

y 4.

5.

[Ans. B] v n ρ

os θ … y n ρ … y now y’ t nθ … Equating equations (i) and (ii) and using equation (iii) in equation (ii), we get y os θ= os θ

[Ans. B] y x x p n nt rv l x y x x y x x Value is in between 20 and 30 So it is 25 [Ans. C] x x gv n t x os t sn t x n x sn t os t t x | t x

os t

y= .x Which is equation of a parabola. 3.

[Ans. A] A.E. D= 1+ 10i C.F = (A cos10 x + B sin 10 x) x

4.

[Ans. C]

5.

[Ans. C]

x

sn t

th

th

th


GATE QUESTION BANK

6.

[Ans. A] Given

Hyperbolic if El ps Compare the given differential equation with standard from A = 1, B = 0, C = 0

= 1 + y2

Integrating ∫

=∫ x

Or t n y = x + c Or y = t n x 7.

Parabolic

[Ans. C] y y x Auxiliary equation, m + 1 = 0 m= 1 C.F =

10.

y

[Ans. C]

The solution for the differential equation is y x Now, y and y , placing these values We get, and y

s nx

s nx

os x

So, y os x s n x or m x m y s nx os x s nx os x x y os x s n x y or x m xm y m x os sn

y

√ 11.

[Ans. A] Given partial differential equation is x

± os x

ẏ ẏ

y

9.

[Ans. D] y t ÿ t

y y

y=

8.

Mathematics

√

√

√

[Ans. D] By back tracking, from option (D) y |x| x or x x = x or x Integrating y ∫ ∫ x x or x x

t

∫ x x or x

x t We know that

x

y (x y

or x

)

or x

is said to be Parabolic if

th

th

th


GATE QUESTION BANK

Mathematics

Complex Variables ME – 2007 1. If x y and (x, y) are functions with continuous second derivatives, then x y + i (x, y) can be expressed as an analytic function of x + i (i = √ ), when (A)

ME – 2014 6.

The argument of the complex number where i = √ π π 2

7. +

(D)

+

(A) 2πi (B) 4πi

An analytic function of a complex variable z x + i y is expressed as z u x y +iv x y where i √ f u(x,y)= 2xy, then v(x,y) must be (A) x + y + onst nt (B) x y + onst nt (C) x + y + onst nt (D) x y + onst nt

9.

An analytic function of a complex variable z = x + i y is expressed as f(z) = u(x, y) + i v(x, y), where i = √ . If u (x, y) = x – y , then expression for v(x, y) in terms of x, y and a general constant c would be (A) xy + (C) 2xy +

ME – 2009 3. An analytic function of a complex variable z = x + iy is expressed as f(z) = u(x, y) +iv(x, y) where i = 1 . If u = xy, the expression of v should be

x

2

2

y 2

2

 k

y (C) (D)

2

 x2 2

 k

 x  y 2  k 2

ME – 2010 4. The modulus of the complex number

(B)

) is

(A) 5 (B) √

(C) 1/√ (D) 1/5

traversed in

8.

(C) 2πi (D) 0

 x  y 2  k

is evaluate

counter clock wise direction. The integral is equal to π (A) 0 – 2 π π – 4 4

+

ME – 2008 2. The integral ∮ z z evaluated around the unit circle on the complex plane for z is

(

x y

along the circle x + y

(C)

(B)

π 2 π

The integral ∮ y x

(B)

(A)

, is

10.

+

(D)

+

If z is a complex variable, the value of is

∫

(A) i (B) 0.511+1.57i (C) i (D) 0.511+1.57i

ME – 2011 5. The product of two complex numbers 1 + i and 2 – 5i is (A) 7 – 3i (C) 3 – 4i (B) 3 – 4i (D) 7 + 3i th

th

th


GATE QUESTION BANK

CE – 2005 1. Which one of the following is NOT true for complex number and ? (A) (B) | (C) | (D) |

2.

̅̅̅̅

=|

|

|≤| |+| | |≤| | | | | +| | 2| | + 2| |

+ +

CE – 2011 6. For an analytic function, f(x + iy) = u(x, y)+iv(x, y), u is given by u = 3x 3y . The expression for v considering K to be a constant is (C) 6x 6y+k (A) 3y 3x + k (D) 6xy +k (B) 6y – 6x + k CE – 2014

Consider likely applicable of u hy’s integral theorem to evaluate the following integral counter clockwise around the unit circle c. ∮s

z z

7.

z

πn

2

+

i i

ECE – 2006 1. The value

of

∮|

2.

-

π/2: singul riti s s t { nπ n 2 } (D) None of the above

∮

dz is

(A)

4πi

(C)

(B)

πi

(D) 1

πi

(C) (D)

the

+

contour

i i

integral

z in positive sense is

|

(A)

(C)

(B)

(D)

For the function of a complex variable W = In Z (where, W = u + jv and Z = x + jy), the u = constant lines get mapped in Z-plane as (A) set of radial straight line (B) set of concentric circles (C) set of confocal hyperbolas (D) set of confocal ellipses

(C) I

CE – 2006 3. Using Cauchy’s is integral theorem, the value of the integral (integration being taken in counter clockwise direction)

can be expressed as

(A) (B)

z being a complex variable. The value of I will be (A) I = 0: singularities set = ϕ (B) I = 0: singularities set =,

Mathematics

ECE – 2007 3. If the semi-circular contour D of radius 2 is as shown in the figure, then the value of the integral ∮

is

j

CE – 2009 4.

The analytic function f(z) = singularities at (A) 1 and 1 (B) 1 and i

5.

has

j2

(C) 1 and i (D) i and i

The value of the integral ∫

j2

2

dz (A) jπ (B) jπ

(where C is a closed curve given by |z| = 1) is (A) –πi (C) (B) (D) πi th

th

(C) π (D) π

th


GATE QUESTION BANK

ECE – 2008 4. The residue of the function

f z 

5.

1

z  2 z  22 2

(A) (B)

at z=2 is

(A)

(C)

(B)

(D)

(C) (D) 2

∮(

) z is

(A) 4π (B) 4π

The equation sin(z)=10 has (A) no real or complex solution (B) exactly two distinct complex solutions (C) a unique solution (D) an infinite number of complex solutions

12.

+ j2 j2

(C) 4π (D) 4π

+ j2 j2

The real part of an analytic function z where z x + jy is given by cos(𝑥). The imaginary part of z is (A) os x (C) sin x (B) sin x (D) sin x

EE – 2007

If f(z) =

+

is given by (A) 2π (B) 2π +

z

, then ∮

z

(C) 2πj (D) 2πj

z and 1 and

(C) (D)

The value of

+

at its poles are

(A)

1.

where C is the

∮

contour |z-i/2| = 1 is (A) 2πi (C) t n z (B) π (D) πi t n z

ECE – 2010 7. The residues of a complex function

(B)

2

ECE – 2014 11. C is a closed path in the z-plane given by |z|=3. The value of the integral

ECE – 2009 6.

Mathematics

EE – 2011 2. A point z has been plotted in the complex plane, as shown in figure below. nit ir l

and z

and

ECE – 2011 8.

The value of the integral ∮

z

ECE\EE\IN – 2012 9. If x = √ then the value of x is ⁄ (C) x (A) ⁄ (D) 1 (B) 10.

Given f (z)

nit ir l

lm

where is the circle |z| is given by (A) 0 (C) 4/5 (B) 1/10 (D) 1

lm

nit ir l

lm

nit ir l

y y lmlm

. If C is a

nit ir l

y

counterclockwise path in the z – plane such that |z+1| =1, the value of ∮

y

z z is th

th

th


GATE QUESTION BANK

EE – 2013 3.

z evaluated anticlockwise around

∮

the circle |z (A) 4π (B) 4.

i|

2 where i √ (C) 2 + π (D) 2 +2i

Square roots of – i, where i = √ (A) i, i (B)

os (

) + i sin (

, is

, are

)

IN – 2005 1. Consider the circle | | 2 in the complex plane (x, y) with z = x + iy. The minimum distant form the origin to the circle is (C) √ 4 (A) √2 2 (B) √ 4 (D) √2 2.

Let ̅, where z is a complex number not equal to zero. The z is a solution of (C) z (A) z (D) z (B) z

os ( ) + i sin ( ) (C)

os ( ) + i sin ( ) os ( ) + i sin ( )

(D) os ( ) + i sin ( os (

)

) + i sin ( )

EE – 2014 5. Let S be the set of points in the complex plane corresponding to the unit circle. {z: |z| } . Consider the (That is, function f(z)=zz* where z* denotes the complex conjugate of z. The f(z) maps S to which one of the following in the complex plane (A) unit circle (B) horizontal axis line segment from origin to (1, 0) (C) the point (1, 0) (D) the entire horizontal axis

IN – 2006 3. The value of the integral of the complex function 3s  4 f(s) (s  1)(s  2) Along the path s  3is (A) 2j (B) 4j

7.

All the values of the multi-valued complex function , where i √ are (A) purely imaginary. (B) real and non-negative. (C) on the unit circle. (D) equal in real and imaginary parts. Integration of the complex function z

, in the counter clockwise

(C) 6j (D) 8j

IN – 2007 4.

For the function

of a complex variable

z, the point z=0 is (A) a pole of order 3 (B) a pole of order 2 (C) a pole of order 1 (D) not a singularity 5.

6.

Mathematics

Let j = √ (A) √j (B) 1

.Then one value of (C)

is

(D)

IN – 2008 6. A complex variable x+j has its real part x varying in the range to + . Which one of the following is the locus (shown in thick lines) of 1/Z in the complex plane?

direction, around |z 1| = 1, is (A) πi (C) πi (B) (D) 2πi

th

th

th


l xis

(Note:

j m gin ry xis

m gin ry xis

j

l xis

IN – 2009 The value of ∮

where the contour

of integration is a simple closed curve around the origin, is (A) 0 (C) (D) (B) 2πj 8.

If z = x+jy, where x and y are real. The value of | | is (A) 1 (C) (D) (B) √

9.

One of the roots of the equation 𝑥 =j, where j is positive square root of 1, is √ (A) j (C) j +j

(D)

√

)

x

j

√

√ y

pl n

l xis

(B)

z is.

∮

l xis

j

7.

Mathematics

IN – 2010 10. The contour C in the adjoining figure is described by x + y . The value of

m gin ry xis

m gin ry xis

GATE QUESTION BANK

(A) 2πj (B) 2πj

(C) 4πj (D) 4πj

IN – 2011 11. The contour integral ∮ / with C as the counter-clockwise unit circle in the zplane is equal to (A) 0 (C) 2π√ (B) 2π (D)

j

th

th

th


GATE QUESTION BANK

Mathematics


2.

4. [Ans. B] By definition C-R equation holds [Ans. A] f(z)=

has simple pole at z = 0

Residue of f(z) at z = 0 lim z z lim os z ∫ z z 2πi (residue at z = 0) 2πi 2πi 3.

[Ans. B] + 4i + 2i 2i + 2i + i + 4i +4 Modulus = √

[Ans. C] Given u=xy For analytic function u v  x y and

[Ans. A] +i 2 2 i + 2i

6.

dw u v  i dz x x 

7.

∫y x

∮

Replacing x by z and y by 0, we get

∮ 8.

+ 2i + i i

r os x

r sin

r sin

r os

r

r

2π

π 2

[Ans. C] u v x y v 2y y 2y + x v 2 y + x v u v y x 2x x 2x + x 2 x x

z2 C 2 Where C is a constant, z v m0 i + 1 2 Integrating, w  i

 (x2  y2  2ixy)   mi  2   or v 

i

x y

y = r sin x y r os

dw  y  ix dz

Where, z = x + iy dw = izdz

i

[Ans. C]

u u i x y

dw  0  iz dz

i

[Ans. C] +i +i i i +i 2i i + i 2 +i rg ( ) t n ( ) i π⁄ 2

u v  y x

By Milne Thomson method Let w = u + iv

or

5.

+ 2i

y 2  x2 2

th

th

th


GATE QUESTION BANK

v

y y

9.

10.

[Ans. B] z ∫ ln z| z

[Ans. B] ∫s

z os z The poles are at z = n + /2 π = π/2 π/2 + π/2 None of these poles lie inside the unit circle |z| =1 Hence, sum of residues at poles = 0 Singularities set = ϕ and 2πi [sum o r si u s o t z t th poles] 2 πi

[Ans. C] iv n u x y v v v x+ y x y v u v u y x x y u u v x+ y y x 2y x + 2x y rm ont ing y t rms only llow v 2 xy +

3.

z z

ln i

ln

ln + ln i ln ln z os z i z i ln i ln z π i ( 2

ln + ln i + i sin i sin π/2

=

=| ̅̅̅̅

z

z= ∮

pplying z z

(

)

u hy’s int gr l th or m, using i .2πi ( )/

/

i

2πi

Now, ∮

/

ln

z

∮

i.e. ∮

)

[Ans. C] (A) is true since ̅̅̅̅

∫

[Ans. A] u hy’s int gr l th or m is f(a) =

+

CE 1.

2.

x x + onst nt

Mathematics

i .2πi 0( )

1/

i .2πi 0( )

1/

2π

̅̅̅̅

4πi

2π

|

(B) is true by triangle inequality of complex number (C) is not true since | |≥| | | | (D) is true since | + |2 = ( + ) ̅̅̅̅̅̅̅̅̅̅̅̅ + = ( + ) (z̅ + z̅ ) = z̅ + z̅ + z̅ + z̅ i ̅̅̅̅̅̅̅̅̅̅̅̅ And | |2 = ( + )

i

4.

4πi

[Ans. D] z z z z + z z i z+i The singularities are at z = i and –i

z

5.

[Ans. C]

= ( + ) (z̅ z̅ ) = z̅ + z̅ z̅ + z̅ ii Adding (i) and (ii) we get | + |2 + | |2 = 2 z̅ + 2 z̅ = 2| | + 2| |

r

∫

os 2πz 2z z *

2

th

th

∫

+ *z

+

th


GATE QUESTION BANK

in

z

is point with in |z|=1(the

los urv w n us integral theorem and say that

7.

[Ans. B] 2 i z +i Multiplying by conjugates 2 i i +i i 2i + i 2 + + i 2 + i

u hy’s

os 2πz [2πi ( )] wh r z 2 2 z [Notice that f(z) is analytic on all points inside |z| ] 2

[2πi

os 2

π

(

/2 )

Mathematics

]

2πi

6.

[Ans. D] f = u + iv u = 3x2 – 3y2 For f to be analysis, we have CauchyRiemann conditions, u v i x y u v ii y x From (i) we have u v x x y ∫ v

ECE 1.

Given ,

∮

I

2.

x +

x

z

z +4 j| 2

|z

2j   2j  2j 2

[Ans. B] iv n

log

1 y  u  iv  loge  x  iy   log x2  y2  i tan1   2 x Since, u is constant, therefore



x v + x 2 i.e. v = 3x2 + f(x) iii Now applying equation (iii) we get u v y x [ x+

1 1  z  4 z  2jz  2j 2

Pole (0, 2) lies inside the circle |z j|=2 y u hy’s nt gr l ormul

∫ x y

y

[Ans. D]



1 log x2  y2  c 2





x +y Which is represented set of concentric circles.

x

y

3.

[Ans. A] s

∮

y x x By integrating, f(x) = 6yx – 3x2 + K Substitute in equation (iii) v= 3x2 + 6yx – 3x2 + K v yx + K

2πj sum o r si u

Singular points are s = Only s= +1 lies inside the given contour lims  1 f s Residues at s= +1 = S1

lims  1 S1

n

th

∮

th



1 1  S 1 2

s

2



s

th

2πj ( ) 2

πj


GATE QUESTION BANK

4.

[Ans. A] Residue of z=2 is

Mathematics

z

d 2 z  2 f z   z2 dz  d 1  2 1   lim lim   2 3 z2 dz  z  2  z  2 32   z  2 

z+

)

z

+

lim

5.

+

(z

+ F(z 2πj 7.

+

)

[Ans. C] X(z) =

[Ans. D] sin z

Poles are Z= 0, Z =1, Z=2 Residue at Z=0 is lim

2i

Residue at Z =1 is lim

2 i

Residue z =2 is lim

2 i (

)

2 i

ut m m

8.

[Ans. A] z+4 ∮ z + 4z +

2 im 2 i

m 2 i

m

i

iz

log

4

2 2 i + i√ 2

√ 2 √

i

2 i

√

z

i

9.

√

i

i

i

√

log i + log( √ π iz log + i ( 2nπ) 2 +log √ π iz i ( 2nπ) + log 2 π z ( 2nπ) ilog( 2

i

)

[Ans. D] f(z) = + + z z z

∮ ∮(

+

log y i log i π i i 2

x log x i log π 2

⟹y √ √

)

10.

z

z+

z F z 2 π j r si u o Residue at z = 0 ( 2- order )

[Ans. C] z z

∫ 2πj

∫

z

*∫

∫

z+

z

z+

where f (z) =1

11. + z

x

log y

z +

x

√ ty

⟹ log y

( 2 infinite number of complex solutions sin z has infinite no. of complex solutions 6.

z + 4z + z+2 + 2 j will be outside the unit circle o th t int gr tion v lu is ‘z ro’

[Ans. A]

i√

i

iz

z

z

[Ans. C] s z lim

2j z + 2j

4+ j

2πj[ 4 + j

)

th

th

4π

th

+ 2j


GATE QUESTION BANK

12.

EE 1.

[Ans. B] Suppose that z u x y + iv x y is analytic then, u and v satisfy the Cauchy Riemann equation u v u v n x y y x r u xy os x u v sin x x y u v os x y x v sin x

z

z +

∮

z

∮ 2 πi 2.

√ + o / is outside the unit circle is IV quadrant 3.

[Ans. A] z 4 ∮ z +4 |z i| 2 z +4 z 4 z 2i For z 2i Residue at z +2i 4 4 +2i z + 2i +4i t z 2i li insi tz 2i li outsi z 4 o∮ 2πi sum o r si u z +4 2πi 2i 4π

4.

[Ans. B] Let + i √ i Squaring both sides we get +2 i i Equating real and imaginary parts

[Ans. B] Pole (z=i) lies inside the circle. |z-i/2|=1. Hence ∮

z+i z

i

2 πi i , wh r

z z

-

π

2i

[Ans. D] Let + i Since Z is shown inside the unit circle in I quadrant, a and B are both +ve and +

√ ow

2

+ i i

+ Since

Mathematics

+

+

wh n

i

2

2

i √2

+

√

wh n

+ o

+i in

qu

r nt wh n

| | √ in

√(

) +(

+

+

√ √

+

i

i

√2 √2 i i +i ( ) √2 √2 i i

√2

+

i √2

√2 √2 i i i +i + i( ) + √2 √2 √2 √2 i +( ) √2 √2 π π os ( ) + i sin ( ) 4 4

)

+

+

th

th

th


GATE QUESTION BANK

or

x π π os ( ) + i sin ( ) 4 4

5.

[Ans. C] z zz

} n s {z: |z| z All point of s will be mapped on the point (1, 0) 6.

[Ans. B] z log z i log z r l n Non-negative

7.

[Ans. C] ∫ x x

)

lim z

IN 1.

2.

int gr tion

2πi

z+

2

2 √2

o |z | king z |z| |z| z

2

uis o th ir l y 4

2

√2

[Ans. C] z z̅ Multiply both the side by z, we get z z̅ z |z| z |z| |z| wh r is ngl o z |z| since is a real quantity so in order to satisfy above equation has to be real quantity = 1 and , (where n = +2 )

z z

r +

√2 x

|

z lim z+ quir

x

2πi r s (f(a)) where a is a

singularity in contour c |z | r n pol s o z z nly z li s insi |z s(

y x x

|z|

Mathematics

π/2 ⁄

z 3.

[Ans. C]

πi

X X -2 -1 Cx y y (Cx ( -3

[Ans. A] | + i | 2 Radius of the circle is 2 and centre is at + i

3 Cx

y(n) n n y(n) )y(n)) 3s  4 1 2 C3 = F(s)  C3  . CC3 (sC 1)(s  2) s  1 s  2 y(n) 3 3 y(n) dz y(n)

2 + i

By Formula, y y ( ( Since, both n n contour, ) )

xy

 z  a  2j

the poles are enclosed by

therefore Value of integral=2πj + 2 2πj For distance to be min. The point P will be on the line passes through origin and centre of the circle. Slope of line OP = Slope of line OC

4.

πj

[Ans. B] Expand by Laurent series

𝑥 th

th

th


GATE QUESTION BANK

5.

[Ans. D]

10. ⁄

tx

j

⁄

(

⁄

log

log

)

2πj 11.

2πj

2

+

[sum o r si

2 o pol

4πj

j

x +

x +

j

∮

⁄

z

∮( + + + + ) z z 2z z The only pole of z is at z , which lies within |z| ∫ z z 2πi (residue) Note: Residue of z at z is coefficient ⁄ of z i.e. 1, here.

x j x + x

j j x +

lim { x +

j ption

⟹z j j ⟹ 2[ j

[Ans. C] z

x+j

|

z

⁄

[Ans. B] x+j

|

z

∮

)

/

x

7.

⁄

log (

π j 2 π j j 2

log

6.

⁄

log (

z=∮

Pole z j Residue at z

⁄

log x

[Ans. D] ∮

)

Mathematics

s tis y th

ov

}

on itions

[Ans. A] u hy’s int gr l ormul is ∫ Here a = 0, then f(0) = sin 0 = 0

8.

[Ans. D] z x + iy p | |= | = |

9.

| |=

|

|=

[Ans. B] Given x3 = j = e+jπ/2 x

⁄

x

os

π

+ j sin

π

√ +j 2 2

th

th

th


GATE QUESTION BANK

Mathematics

Laplace Transform ME – 2007 1. If F (s) is the Laplace transform of function f (t), then Laplace transform of t

 f (t) dt is 0

(A)

F (s)

(C) sF (s) – f (0)

(B)

F (s) f (0)

(D) CE – 2009 1. Laplace transfrm of the function f(x) = cosh(ax) is (A) (C)

ME – 2009 2.

The inverse Laplace transform of is (A) (B)



1 s s 2



(B) (C) 1 – (D)

ME – 2010 3. The Laplace transform of a function . The function

is

is

(A) (B)

(C) (D)

ME – 2012 4. The inverse Laplace transform of the function F(s)

is given by

(A) (B)

(C) (D)

(D)

ECE - 2005 1. In what range should Re(s) remain so that the Laplace transform of the function exists. (A) (C) (B) (D) ECE – 2006 2. A solution for the differential equation x’(t)+2x(t)= (t) with initial condition x( )=0 is (C) (A) (D) (B) ECE – 2008

ME – 2013 5. The function equation

3.

Consider the matrix P = *

satisfies the differential

value of eP is

and the auxiliary

conditions,

+ . The

(A) *

+

(B) [

]

. The

Laplace transform of

is given by

(A)

(C)

(B)

(D)

ME – 2014 6. Laplace transform of The Laplace transform of

(C) [

]

(D) [ is

.

]

ECE - 2010 4. The trigonometric Fourier series for the waveform f(t) shown below contains th

th

th


GATE QUESTION BANK

Mathematics

ECE – 2013 9. A system is described by the differential equation

=x(t). Let x(t)

be a rectangular pulse given by , Assuming that y(0) = 0 and (A) Only cosine terms and zero value for the dc component (B) Only cosine terms and a positive value for the dc component (C) Only cosine terms and a negative value for the dc component (D) Only sine terms and a negative value for the dc component. 5.

the Laplace transform of y(t) is

Given [

10.

]

then the value of K is (A) 1 (C) 3 (B) 2 (D) 4 ECE– 2011 6.

[

If

]

then the initial

and final values of f(t) are respectively (A) 0, 2 (C) 0, 2/7 (B) 2, 0 (D) 2/7, 0

The maximum value of the solution y(t) of the differential equation y(t) + ̈ with initial condition ̇ and ≥ (A) 1 (C) (B) 2 (D) √

ECE – 2014 11. The unilateral Laplace transform of . Which one of the following is the unilateral Laplace transform of ?

ECE/EE/IN – 2012 7. The unilateral Laplace transform of f(t) is . The unilateral Laplace transform

8.

of t f(t) is (A) –

(C)

(B)

(D)

Consider the differential equation

|

|

The numerical value of (A) (B)

EE – 2005 12. For the equation (t) + 3 (t) + 2x(t) = 5, the solution x(t) approaches which of the following values as t ? (A) 0 (C) 5 (D) 10 (B) EE – 2014

|

is

13.

(C) (D)

Let

be

the

transform of signal x(t). Then, (A) 0 (C) 5 (B) 3 (D) 21 th

th

th

Laplace is


GATE QUESTION BANK

14.

Mathematics

[ Let g: [ be a function [ ] where [x] defined by g(x) represents the integer part of x. (That is, it is the largest integer which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is_______


5.

[Ans. C]

[Ans. A] From definition, We know ∫

2.

Taking Laplace transformation on both sides [ ] [ ] ( ) ( )

[Ans. C]

1 1 1 1    (s  s) s(s  1) s (s  1) 2

(

3.

)

( )

(

[

)

(

) (

[

[Ans. A] [

]

6.

4.

)

]

s and constant

]

[Ans. B] It is the standard result that L (cosh at) =

ECE 1.

[Ans. A] [

[Ans. D] {

]

[Ans. D]

CE 1.

[

)

]

[

Matching coefficient of in numerator we get,

(

]

} 2. {

[Ans. A] ̇ (t) + 2x (t) = (t) Taking Laplace transform of both sides , we get

}

th

th

th


GATE QUESTION BANK

sX(s) X(0) + 2X(s) = 1 1 X(s) = s2 From Inverse Laplace transform gives, we get [ ] 3.

[Ans. D ] eP= [

Mathematics

cosine terms and a negative value of the dc component. 5.

[Ans. D]

[

]

] [

0 1  and P=   2  3 s 1  Where =   2 s+3 s  3 1  1  s  1s  2 2 s s 3    s  1 s  2 =   2  s  1 s  2  

]

[

1

1  s  1 s  2      s s  1s  2 

6.

]

[Ans. B]

Using initial value theorem:

 eP

 2 1  s  1  s  2   2  2  s  1 s  2

=

1 1   s  1 s  2    2 1   s  2 s  1  

=[

⁄

] =2

4.

[Ans. C] Since f(t) is an even function, its trigonometric Fourier series contains only cosine terms

∫

7.

∫

*∫

∫

[Ans. D]

+ t

[

(

)] 8.

[

[Ans. D]

]

Therefore, the trigonometric Fourier series for the waveform f(t) contains only

Taking Laplace transform on both the sides. We have, th

th

th


GATE QUESTION BANK

11.

Mathematics

[Ans. D]

By Laplace transform property, [

]

[

]

(

[

] [

|

] 12.

9.

[Ans. B] =5 By taking Laplace transform

[Ans. B] Writing in terms of Laplace transform

( ⁄

X(s) = (

) 13.

[Ans. B]

(

)

( ( 10.

)

[

)

]

(

)

)

[Ans. D] 14.

[Ans. 0.5]

∫ For t =

∫

|

Value of constant term = 0.5 + sin

√

th

th

th


GATE QUESTION BANK

Network Theory

Network Solution Methodology ECE - 2006 1. A negative resistance R is connected to a passive network N having driving point impedance (s) as shown below. For (s) to be positive real, Rneg

4.

In the interconnection of ideal source shown in the figure, it is known that the 60V source is absorbing power. 20 V + I

60 V 12 A

N

Z2(s)

Which of the following can be the value of the current source I? (A) 10 A (C) 15A (B) 13A (D) 18A

Z1(s)

(j )

(A) |R

|

Re

(B) |R

|

| (j )|

(C) |R

|

(D) |R

|

m

ECE - 2010 5. In the circuit shown, the power supplied by the voltage source is

(j ) (j )

ECE - 2007 2. For the circuit shown in the figure, the Thevenin voltage and resistance looking into X –Y are

1Ω 1Ω

1Ω

10V 1

1Ω

2

1Ω

1Ω

i 2i 1Ω

2

(A) 4/3 V 2Ω (B) 4V 2/3 Ω

2Ω

(C) 4/3 V 2/3Ω (D) 4V 2Ω

ECE - 2009 3. In the circuit shown, what value of RL maximizes the power delivered to RL? V 4Ω

4Ω V

(A) 0 W (B) 5 W

(C) 10 W (D) 100 W

ECE - 2011 6. In the circuit shown below, the value of R such that the power transferred to R is maximum is 10 Ω

10 Ω

R

10 Ω

4Ω

V 100V

(A) 2.4 Ω (B) 8⁄3 Ω

1Ω

R

5V

(C) 4 Ω (D) 6 Ω

1

2V

(A) 5Ω (B) 10Ω th

th

(C) 15Ω (D) 20Ω th


GATE QUESTION BANK

7.

In the circuit shown below, the current I is equal to

11.

VB =6 V, then VC R

j4Ω

j4Ω

If VA

Network Theory

V

R

R

2Ω

VD is R

V

R

R

R

~

1Ω

6Ω

14 0 V

R V

5V

6Ω

(C) 2.8 0 A (D) 3.2 0 A

In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

25Ω

(A) 6.4 – j4.8 (B) 6.56 – j7.87

Q

ECE/IN - 2012 9. The average power delivered to an impedance (4 j3) by a current 5cos (100t+100) A is (A) 44.2 W (C) 62.5 W (B) 50 W (D) 125 W

RC

RB RA

(C) 1/k

(A) k (B) k 13.

ECE/EE/IN - 2012 10. The impedance looking into nodes 1 and 2 in the given circuit is i

9kΩ

Rc

Rb

(C) 10 + j0 (D) 16 + j0

99i

1kΩ

(C) 3 V (D) 6 V

Ra

j50Ω

15Ω

V

ECE/EE/IN - 2013 12. Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection are scaled by a factor k, k>0, the elements of the corresponding star equivalent will be scaled by a factor of

P

j30Ω 16 0

2

(A) 5 V (B) 2 V

6Ω

(A) 1.4 0 A (B) 2.0 0 A 8.

10V

(D) √k

Three capacitors C1, C2 and C3 whose values are 10μF 5μF and 2μF +respectively, have breakdown voltages of 10V, 5V and 2V respectively. For the interconnection shown below , the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in μC stored in the effective capacitance across the terminals are respectively, C2 C3

1 100Ω 2

(A) 50  (B) 100 

C1

(C) 5 k (D) 10.1 k

(A) 2.8 and 36 (B) 7 and 119 th

th

(C) 2.8 and 32 (D) 7 and 80 th


GATE QUESTION BANK

ECE - 2014 14. For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance of the first section to the input impedance of the second section is (A) (C) (B) (D)

18.

15.

19.

Consider the configuration shown in the figure which is a portion of a larger electrical network

Network Theory

In the figure shown, the value of the current I (in Amperes) is______. 5Ω 5Ω

5V

10Ω

1

In the circuit shown in the figure, the value of node voltage V is 10 0

i i

V

V 4Ω

R R

i

i

i

For R 1Ω and currents i =2 i 1 i 4 which one of the following is TRUE? (A) i 5 (B) i 4 (C) Data is sufficient to conclude that the supposed currents are impossible (D) Data is insufficient to identify the currents i i and i

17.

j6Ω

6Ω

j3Ω

R

i

16.

4 0

A Y-network has resistances of 10Ω each in two of its arms, while the third arm has a resistance of 11 Ω. In the equivalent -network, the lowest value (in Ω.) among the three resistances is ________

(A) 22 + j 2 V (B) 2 + j 22 V 20.

(C) 22 – j 2 V (D) 2 – j 22 V

The circuit shown in the figure, the angular frequency (in rad/s) at which the Norton equivalent impedance as seen from terminals b-b′ is purely resistive is_____________. 1Ω

1F b

10 cos t (volts)

~

0.5

b′

21.

Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance (A) in series with a current source (B) in parallel with a voltage source (C) in series with a voltage source (D) in parallel with a current source

For the Y-network shown in the figure, the value of R (in Ω) in the equivalent ∆-network is ____. R

5Ω

3Ω 7.5Ω

th

th

th


GATE QUESTION BANK

22.

The magnitude of current (in mA) through the resistor R in the figure shown is__________ R 1kΩ R

10 m

R

2kΩ

R

23.

2m

4kΩ

3kΩ

The equivalent resistance in the infinite ladder network shown in the figure, is R . 2R R

R

R

R

R

R

R

(A) Very low (B) L/3

(C) 3L (D) Very high

EE - 2007 2. A 3V dc supply with an internal resistance of 2Ω supplies a passive non-linear resistance characterized by the relation VNL = . The power dissipated in the non-linear resistance is (A) 1.0 W (C) 2.5 W (B) 1.5 W (D) 3.0 W EE - 2008 3. In the circuit shown in the figure, the value of the current i will be given by 1Ω 1Ω

R

The value of R /R is________ 24.

Network Theory

5V

a + V

1Ω 1Ω

3Ω 1Ω

b

4V

1Ω 1Ω

i

The circuit shown in the figure represents a (A) 0.31 A (B) 1.25 A

R

4.

Assuming ideal elements in the circuit shown below, the voltage Vab will be 2Ω

a

(A) voltage controlled voltage source (B) voltage controlled current source (C) current controlled current source (D) current controlled voltage source EE - 2006 1. The three limbed non ideal core shown in the figure has three winding with nominal inductances L each when measured individually with a signal phase AC source. The inductance of the winding as connected will be R

(C) 1.75 A (D) 2.5 A

+ 1A

Vab

+

i

5V

b

(A) – 3 V (B) 0 V

(C) 3 V (D) 5 V

EE - 2009 Statements for Linked Answer Questions 5 & 6: 2kΩ 5V

+

+

3VAB A

2kΩ

1kΩ B

th

th

th


GATE QUESTION BANK

5.

For the circuit given above, Thevenin’s resistance across terminals A and B is (A) 0.5kΩ (C) 1kΩ (B) 0.2kΩ (D) 0.11kΩ

the the

Network Theory

EE - 2010 10. If the 12 Ω resistor draws a current of 1A as shown in the figure, the value of resistance R is 1Ω

6.

7.

8.

For the circuit given above, the Thevenin’s voltage across the terminals and B is (A) 1.25V (C) 1V (B) 0.25V (D) 0.5V How many 200W/220V incandescent lamps connected in series would consume the same total power as a single 100W/220V incandescent lamp? (A) not possible (C) 3 (B) 4 (D) 2

(A) (B) (C) (D)

2Ω

Is = 5A

4V

(5A; Put Vs=20V) (2A; Put Vs =8V) (5A; Put Is = 10A) (7A; Put Is= 12A)

11.

1kΩ 6V

(A) 0mA (B) 1mA

(C) 2mA (D) 6mA

6V

(C) 8Ω (D) 18Ω

As shown in the figure, a 1Ω resistance is connected across a source that has a load line v + i = 100. The current through the resistance is i Source

1Ω

V

(A) 25A (B) 50A

(C) 100A (D) 200A

EE - 2011 12. In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3Ω is R

10 V

The current through the 2 kΩ resistance in the circuit shown is C 1kΩ 1kΩ 2kΩ 1kΩ

12Ω

1

(A) 4Ω (B) 6Ω

For the circuit shown, find out the current flowing through the 2Ω resistance. Also identify the changes to be made to double the current through the 2Ω resistance.

V

9.

2

R

6Ω

(A) Zero (B) 6Ω

3Ω

oad

(C) 3Ω (D) Inifnity

EC/EE/IN - 2012 13. Assuming both the voltage sources are in phase the value of R for which maximum power is transferred from circuit A to circuit B is

th

th

th


GATE QUESTION BANK 2Ω

~

16.

R

j1Ω

10V

Circuit

~

3V

In the figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The current I is______

Circuit

(A) 0.8 Ω (B) 1.4 Ω

Network Theory

300V

R

(C) 2 Ω (D) 2.8 Ω

EC/IN/EE - 2013 14. The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage V 100V is applied across WX to get an open circuit voltage V across YZ. Next , an ac voltage V = 100V is applied across YZ to get an open circuit voltage V across WX. Then,V /V V /V are respectively. W

17.

Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 μH and 240 μH. Their mutual inductance in μH is_______

18.

The total power dissipated in the circuit, shown in the figure, is 1 kW. 10

~

2

1Ω

R

V

ac source

oad 200V

1:1.25

The voltmeter, across the load, reads 200 V. The value of is__________

Y

19.

The line A to neutral voltage is 10 15° V for a balanced three phase star-connected load with phase sequence ABC. The voltage of line B with respect to line C is given by (A) 10√3 105 V (C) 10√3 75 V (B) 10 105 V (D) 10√3 90 V

20.

The Norton’s equivalent source in amperes as seen into the terminals X and Y is _______

Z

X

(A) 125/100 and 80/100 (B) 100/100 and 80/100 (C) 100/100 and 100/100 (D) 80/100 and 80/100 EE - 2014 15. The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is ___________ 10

2.5V

8

2.5Ω 100V

5Ω

80V

5Ω 15V

5Ω

2

5V

th

th

th


GATE QUESTION BANK

21.

The power delivered by the current source, in the figure, is_____________

3.

Which one of the following equations is valid for the circuit shown below? 1Ω 1Ω I3 I1

1V

1Ω

1Ω

I2

1Ω

1V

Network Theory

+

1Ω

2

1Ω

5V

I5

I6

1Ω

1Ω

1Ω

22.

An incandescent lamp is marked 40 W, 240V. If resistance at room temperature (26°C) is 120 Ω, and temperature coefficient of resistance is 4.5 10 /°C, then its ‘ON’ state filament temperature in °C is approximately___________

IN - 2006 1. The root – mean – square value of a voltage waveform consisting of a superimposition of 2V dc and a 4V peak – to – peak square wave is (A) 2 V (C) √8 V (B) √6 V (D) √12 V

(A) (B) (C) (D) 4.

I

1V

5.

1A

(C) 1A (D) 2A

In the circuit shown below the maximum power that can be transferred to the load is 10√2 sin(1000t)

1Ω

(C) 2.5W (D) 3.0W

10Ω

10 m

i(t)

(A) 0W (B) 1.0W

1Ω

(A) 0A (B) 0.5A

3Ω 3Ω

6Ω

0 0 0 0

The current I supplied by the dc voltage source in the circuit shown below is

IN - 2008 2. The power supplied by the dc voltage source in the circuit shown below is

3V

I4 I7

(A) 250 W (B) 500 W

th

th

(C) 1000 W (D) 2000 W

th


GATE QUESTION BANK

IN - 2009 6. The source network S is connected to the load network L as shown by dashed lines. RL

2Ω

+

+ 10 V

3V

Source Network S

Network Theory

IN - 2014 9. The circuit shown in the figure contains a dependent current source between A and terminals. The Thevenin’s equivalent resistance in kΩ between the terminals C and D is ___________. 5kΩ 5kΩ C

10 V

10 V

Load Network L

The power transferred from S to L would be maximum when RL is (A) 0Ω (C) 0.8 Ω (B) 0.6 Ω (D) 2Ω

V

IN - 2010 7. A 100 Ω , 1W resistor and a 800 Ω , 2W resistor are connected in series. The maximum dc voltage that can be applied continuously to the series circuit without exceeding the power limit of any of the resistors is (A) 90V (C) 45 V (B) 50 V (D) 40V IN - 2011 8. The current I shown in the circuit given below is equal to 10Ω

10 V

(A) 3 A (B) 3.67 A

10Ω

10

10Ω

(C) 6 A (D) 9 A

th

th

th


GATE QUESTION BANK

Network Theory

Answer Keys and Explanations ECE 1.

4. [Ans. A] (s) = (s)

I

(S)

R ( (s))

+

60V

|

I

V x

2i

i 1Ω

2

2Ω

y

For V Apply nodal analysis, V V 2i 2 V 0 V i 2 1 V V 2 2V 2i 0 2 2 2 V V 4V Similarly, 2 V V 4V R 2Ω 3.

[Ans. C] For maximum power transfer, R V 4Ω

4Ω

5.

12 A

[Ans. A] 3 1Ω 1Ω

1Ω (

3)

3 10V

(

2)

1

1Ω

2 (

2)

1Ω Fig .1

R

The current through all the branches are marked as shown in Fig. 1. Apply KVL to outer loop 2( 3) 2( 2) 10 4 10 10 0 Power supplied by 10 V 10 0 0

4Ω

V 100V 100V 100 (100 V ) 8 4 Also V 50V 12.5 12.5 25 V R 4Ω R ⁄

12 A

In the given circuit, the current through the branch of 60 V source is (12 –I) as shown in Fig. The source of 60 V absorbs power, only if P =(12 – I)60 is +ve. i.e., I<12. The value of the current source, I can only be 10A given in option (a), as the currents given in other options are more than 12 A.

[Ans. D] 1Ω

(12 – I)A

0 |R

20 V

12 A

R

(R

m( (s)) R For (s)to be +ve & real, Re ( (s)) R |Re( (s))| 2.

[Ans. A]

V

6.

[Ans. C] 10

10 10

R

15

4Ω Fig. 2

For Maximum power transfer to RL th

th

th


GATE QUESTION BANK

R R R Thevenin’s resistance seen across the terminals of R into the rest of the network. The relevant circuit is shown in Fig.2, where the independent current source is open circuited and the voltage sources are short circuited. R 10 (10 10) R R 15Ω 7.

Network Theory

8.

[Ans. A] The Norton equivalent current is 25 16 0 (40 30) 3 8 tan ( ) 8 36.86 4 (6.4 j4.8)

9.

[Ans. B] The load consists of a resistance and a capacitance of this, only R is passive and consumes power So P = R

[Ans. B] To find the current I in the given circuit in Fig. (1), the delta network with 6Ω each is converted to a star as shown in Fig. (2)

=( )

4

√

14 0 V

~

*Note rms value of

j4 Ω

j4 Ω

10.

6Ω

50 cos t

√

+.

[Ans. A] i

6Ω Fig. 1

99i

1kΩ

6Ω 2Ω

9kΩ

2Ω

100Ω 6Ω

6Ω 2Ω

i

(

99i )

Fig. 2

Then the given circuit reduces to Fig. (3), (4) and (5) (2 j4)||(2 j4) 5Ω and Where 14 0 2 0 7

V

14 0 2Ω

2Ω 2Ω

Fig. 3

5Ω 14 0

14 0

100Ω

V

99i

V

After connecting a voltage source of V V V (10k)( i ) 100( 99i i ) 10000i 100( 100 i ) 100 10000i 20000i 100 100 i ( ) [ ] 20000 200 V 100 99i i

j4Ω

j4Ω

10kΩ

100 [

7Ω

2Ω

R

50 V

100 ( 50

200

)]

50Ω

Fig. 4 th

th

th


GATE QUESTION BANK

11.

[Ans. A] R

R

2Ω

V

R

R

14.

[Ans. C] For maximum power transfer Load impedance = complex conjugate of source impedance

15.

[Ans. A]

R

V

R R

Network Theory

10V

1Ω R V

5V

i

From the given circuit,V V V V 3 2 KC at ‘ ’ gives V V 2 3 0V V 1 12.

13.

[Ans. B] Consider R k R R R R R R R R k

i

V

2

k R k 3k

6V R R

5μF

C

5V

(Safe voltage)

V C

i

5V

10V

k.

2μF

(Safe voltage) (Safe voltage)

For safe warping voltage across should not increased from 2V C V V C C V Total safe voltage across C and C 5μV 2 5μ 2μ 5μV 2 7μ 14 [V 2.8V] 5 C C (C C ) 80 C 10μF (5μ 2μ) μF 7 80 Total charge C . V 2.8 32C 7

2 4 1

i

KC at node i i i KCL at node B i i i KCL at node C i i i Put the value of i and i in equation 2 ( 1) i i 1 Put i and i in equation 1 ( 4) i i 3 Put the value of i and i in equation 3 i 2 i 5

2V

10μF

R

i

[Ans. C] C

i i i

i

16.

[Ans. *] Range 29.08 to 29.10 R 10Ω

10Ω

R

R

11Ω

R 29.09 R

32

R

32

So lowest value is 29.09 Ω th

th

th


GATE QUESTION BANK

17.

[Ans. D] Norton’s theorem: Here for a complex network, after isolating the element, we short circuit the two ends and find the current . is the current of the independent current source with the equivalent resistance in parallel

10 0 V V1

4 0

5Ω

20. 10Ω

5V

3jΩ

6Ω

[Ans. *] Range 1.9 to 2.1 The Norton’s equivalent circuit is 1Ω

1F

Voltage source shorted

5 0.25 mp 20 For current 5Ω

In

5Ω

0.5

doiman 1 j j

(1||0.5j ) 1

1

5 20

0.5j 1 0.5j 0.5j (1 0.5j ) j 1 0.25 0.5j (1 0.5j ) j(1 0.25 ) (1 0.25 ) Equating the imaginary part to zero, we get 0.5 1 0.25 0 0.25 1 2rad/sec

10Ω

0.25 mp

(0.25 0.25) 0.50 mp 19.

6jΩ

Applying KCL to the super node we have V V V 4 0 0 . 3j 6 6j From super node we get V V 10 0 . Solving and for V we get V 2 22j V

[Ans. *] Range 0.49 to 0.51 Apply superposition theorem For voltage 5Ω

V2 4Ω

R

18.

Network Theory

[Ans. D] Using super node concept, we can treat nodes 1 and 2 to gather as a super node as shown in the figure by the dotted lines

th

th

th


GATE QUESTION BANK

21.

R R

[Ans. *] Range 9 to 11 R

24.

Network Theory

2.618

[Ans. C ]

3Ω

5Ω

i

R

i

7.5Ω

5

22.

In the above circuit in the output side there is a dependent current source which is controlled by the input current and hence it is a current control current source.

7.5 3 7.5 7.5 7.5 2 5 3 10 7.5

R

3

5

EE 1.

[Ans. *] Range 2.79 to 2.81 By source transformation theorem 2k 1k 4k

20V

[Ans. A] The inductance of all three coils are ‘ ’ and they are connected to same line carrying the same current and set up the flux in the same direction

8V 2 2

by KV 20 28 10k 23.

10k

3k 8 0

2.8 m All these fluxes linked with each other and they will balance each other. So net flux will reduce drastically. Thus net L will be very low

[Ans. *] Range 2.60 to 2.64 We know that in a infinite ladder network if all resistance are comprises of same value R then the equivalent resistance is (

√ )

. Then the above given network

2.

[Ans. A]

can be redrawn as R series with R equivalent as follows R R R

R

R

1.618R

(1

2Ω 3V

√5)R 2

+ I

V

3 2 1 Power delivered by source = 3×1=3W.

2.618R

th

th

th


GATE QUESTION BANK

Power dissipated by 2Ω resistor 2 2 . Power dissipated in non-linear element = 3-2=1W

6.

Network Theory

[Ans. D] 2kΩ

3V

1kΩ

[Ans. B] 5 1 V 2 Also, 4 V Also, V V

2.5V 4i

V 2

V

To calculate thevenin’s voltage terminals A-B are kept open. Applying source transformation into correct source

4i . (1)

V

1.25 V

5

2

C

4i (From (1))

5

3V

3V

2kΩ

C

2kΩ

2kΩ

1kΩ

Applying source transformation current source is transformed into voltage source.

[Ans. B] Thevenin’s resistance is calculated using the circuit shown in fig. (1) and (2), where independent voltage source is short circuited

1kΩ

C

3V

5 V 2

1kΩ

V

I A

2kΩ

Applying KVL, {I is assumed to be in mA} 5 1k 3V 1k 0 2 5 2k 3V 0 (1) 2 V (2) 1k Put the value of in quation (1) 5 2V 3V 0 2 V 0.5V

V

1kΩ

B Fig. 1 Fig. 1 3 V 10 )

I V

A

10

1kΩ

V

V

1kΩ

(

5.

[Ans. A] V 2i

V)

4(V

4V

4(2.5 1.25) i 1.25 4.

V

2kΩ

5V

3.

C

Fig. 2

B

7.

[Ans. D] For a lamp, P KV For 200 ⁄220V lamp, K 200⁄ 220 Consider n lamps connected in series, Total power consumed n K 110

Write the loop equation : ( V 3V V 10 )10 10 V 5V 10 V 10 R Ω 0.2kΩ 5

100

th

th

n

110

th

100

n 2


GATE QUESTION BANK

8.

[Ans. B] The relevant circuit is shown in fig. As the voltage across 2Ω 4V 4 2 2 In order to double the current through 2Ω resistance, V is to be doubled (Put V 8V) ] Note that the 5 A source has no effect on the answer. However it gives 3A current through the voltage source as shown in fig.

V

5

4V

R

R 6R 6 R 6Ω

3 R 13.

Network Theory

[Ans. A] 2Ω

~

j1Ω

10V

Circuit

2Ω

[Ans. A] As the ABCD bridge is balanced,

0

[Ans. B] Current through R =1A By KVL, 1.R +6=12 R=6

2Ω

2Ω

[Ans. B] V i 100 and V i .1(by ohm’s law) 2i 100  i 50 [Ans. B] To calculated circuited R

~

5V

10 2.5 4 So in our circuit

10V

12.

Circuit

2Ω 10V

11.

3V

5V

Fig.

10.

~

For maximum power transfer Source impedance = Load impedance z R

3

9.

R

~

R

2.5

~

5V

3V

pply KV in loop 5 2.5R 3 0 2 [R 0.8Ω] 2.5

voltage source is short 14.

[Ans. B] W

1:1.25 100 V

6

125 V

Y V

R X

Z

V 6R R 6||R 6 R According to maximum power transfer theorem

100

V 0.8 when V

th

th

V

turns ratio 125 V 100V 100V

th

V


GATE QUESTION BANK 2

18.

[Ans. *] Range 17.3 to 17.4 (10)√

200

V

2′

Thevenin’s circuit seen by 2 2′ will be as follow V 100V nd R 0.2||0.8 negligible V 100V V V turns ratio 100V 1 1.25 80V

R (2) (1)

Power dissipated 4 10R 1000 996 R 10 400 R 400 99.6 17.33Ω

100V

0.8

Network Theory

19.

R

10

[Ans. C] C

120 120

15.

16.

[Ans. 330] Power absorbed by battery 100V = 100 10 1000 Power supplied by battery 80V 80 8 640 Current through 15V battery 10 8 2 Power supplied by battery 15V 15 2 30 Total power absorbed 1000 640 30 330 att

120

V

lags voltage A by 90

V 90 √3 10 Angle 90 with respect to A Hence V

20.

√3 10 ( 90 10√3 75

[Ans. 2] 2.5V

[Ans. 10] 300

(25

600

(50 10

15)

2.5Ω

2 )

) i 5Ω

5Ω 5V

17.

[Ans. 35] Total flux linkage = total

i

Terminal x – y shorted Applying superposition i due to 5V (5) 5 (5 ) 7.5

380μm 240μm 4

140μm 35μ

5

5 20 7.5 i due to 2.5V

th

3

th

5 10

th

0.5


GATE QUESTION BANK

2.5 1 0.5 (2.5 2.5) 2 Total i 0.5 0.5 1 (NOTE: Option not matching with IIT website) 21.

2.

Network Theory

[Ans. D] The given circuit in Fig. 1 is simplified as shown in Fig. 2 and Fig.3

[Ans. 3]

3  6 

V1=3 V

1V

2  1 

1 

V1

Fig.1

1Ω 1V

O

3. Power delivered by current source Applying nodal at V (

) ( )

[Ans. D] The circuit is shown in Fig. marking the nodes: P,Q,R and S Apply KCL at nodes,

V

2

3 3 v 2 Power delivered

2v

22.

I1

1Ω

I2

1Ω

[Ans. *] Range 2470 to 2471

I5

1Ω

I3

1Ω

I6

R

1Ω

1Ω

1Ω

I4

P

I7

S

At node, P . (1) At nodeQ , . . (2) At nodeR , . (3) At nodeS , . (4) From (3) and (4) 0 4.

[Ans. A] I

P I1

[Ans. C] For d.c voltage of 2 V, M.S.V = 4 For square wave voltage with peak – to – peak value of 4 V or amplitude = 2V, M.S.V = 4 + 4 = 8 R. . S. V

Q

5V

3

40 R (240) R 40 R R (1 T) R is resistance of room temperature (240) 120(1 4.5 10 T) 40 11 4.5 10 T 11 1000 T T 4.5 11 1000 T 26 2470.44 4.5 IN 1.

Fig.3

Fig.2

1Ω

2 O

( )

1 

V1

From Fig.3, I = 1 A, power supplied by 3 V d.c source = P = V1 I = 3 1 3 W

1Ω

V

I

+

1V

1Ω

1A

The circuit is shown in Fig. Voltage across 1 =1V 1 = 1A Apply KCL at node, P

√8 V

th

th

th


GATE QUESTION BANK

I+1= I1, or I+1=1 , 0 5.

I +

Source Network S

10Ω

3V +

Load Network L

7 2 R Power supplied by the 10V source,

10 m

P Power dissipated in 2 resistance, 98 P 2 (2 R ) Power transferred to the network,

10√2 Sin (1000 t) 10 /

Phasor of i(t) R j

10√2 10 j10 10 10 10 j 10 The Thevenin equivalent circuit is shown in Fig. 2

P (R )

P

P

(

P would be maximum, if

)

0

70 2 98 0 (2 R ) (2 R ) 196 196 98 70 2 R 2 R 70 35 98 28 R 2 0.8  35 35

⃗ V 7.

[Ans. C] Resistor 1 : 100 Ω 1 Resistor 2 : 800 Ω, 2 W Maximum current that withstand

Fig. 2 10 j 10 ⃗⃗⃗⃗ V 10√2 10√2(10 j 10) 200 e For maximum power transfer to load, 10 j 10 R 10Ω ⃗⃗⃗⃗ V ⃗⃗⃗ 10 e 20 i (t) 10 sin( 45 ) 10 (rms) 10 √2 Power transferred to load, 100 (rms) R P 10 500 2 6.

+ 10 V

10Ω

i(t)

RL

2Ω

[Ans. B] The circuit is shown in Fig. 1

i(t)

Network Theory

√

1 100

resistor can

1 10

Similarly, √

2 800

1 20

If these two resistors are connected in series. 100Ω

800Ω

I

[Ans. C] The circuit is shown in Fig.

V

Then maximum value of th

th

th


GATE QUESTION BANK

V

8.

(100 800) 1 (900) 45Volts 20

9.

[Ans. 20] 5kΩ

[Ans. A] The given circuit is Fig. 1 Convert the (10 V, 10 Ω) voltage source across A, B to the left into a current source, (1 A, 10 Ω). The resultant circuit is shown in Fig. 2

5kΩ

10 V

10 V

C

V

For R calculation Independent voltage source should be short circuited So

10Ω

10 V

10

10Ω

10Ω

5k

Fig. 1

1

Network Theory

10Ω

10Ω

5K

10

C

10 V 1V( ssume) so V 1

10Ω R

So *R

Fig. 2

Apply KCL at Node A 10 V 1 V V 10 5k 5K 1 V .5 V 2V 1.5 V 0.75V 1 V 1 0.75 So 5k 5k 0.25 5k 5 10 0.05m 1 [R 20kΩ] 0.05m

The circuit is further simplified as shown in Fig. 3 and 4 1

5Ω

10

10Ω

Fig. 3

9

5Ω

10Ω

Fig. 4

9

5 15

+

3

th

th

th


GATE QUESTION BANK

Network Theory

Transient/Steady State Analysis of RLC Circuits to DC Input Q

ECE - 2006 1. In the figure shown, assume that all the capacitors are initially uncharged. If Vi(t) =10u(t) Volts, then V0(t) is given by

P

Q2 P2 +

S

S2

0

(t) F

1K +

+ Vi(t)

(A) (B) (C) (D)

4 F

4K

1µF

Assume that the capacitor has zero initial charge. Given that u(t) is a unit step (t) across the function, the voltage capacitor is given by (A) ∑ ( ) tu(t nT) (B) u(t) + 2 ∑ ( ) u(t nT) (C) tu(t) + 2 ∑ ( ) (t nT) u(t nT) ) ) (D) ∑ 0 e ( +0 e ( ]

Vo(t)

8e Volts 8( e ) Volts 8 u(t) Volts 8 Volts

ECE - 2007 2. In the circuit shown, Vc is 0 volts at t = 0 sec. For t>0, the capacitor current iC (t), where t is in seconds, is given by 20k

Common Data for Questions 4 and 5: The following series RLC circuit with zero initial condition is excited by a unit impulse function δ(t)

i

1H

±

20k 10V

(A) (B) (C) (D)

0.50 exp ( 0.25 exp ( 0.50 exp ( 0.25 exp (

+ 4 F -

VC

25t)mA 25t)mA 12.5 t)mA 6.25 t)mA

ECE - 2008 3. The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows For 2nT ≤ t< (2n+ )T, (n = 0, , 2…) S to P and S2 to P2 For (2n+ )T ≤ t< (2n+2)T, (n = 0, , 2…) S to Q and S2 to Q2

δ(t)

4.

Ω

+ 1F

For t > 0, the output voltage (A) (B) (C) (D)

5.

√ √ √ √

(e

√

e

VC(t)

(t) is

)

te e e

cos ( ⁄

√

t)

√

t)

sin (

For t > 0, the voltage across the resistor is (A)

√

(e

(B) e (C) (D) th

√ √ th

√

e

*cos (

√

) )

√

√

)

√

)

e

sin (

e

cos (

th

sin (

√

)+


GATE QUESTION BANK

6.

Network Theory 0Ω

In the following circuit, the switch S is closed at t=0. The rate of change of current

1.5A

(0 ) is given by

15mH

R

i(t)

(A) 0

(C)

(B)

(D)

(A) (B) (C) (D)

L

Rs

(

)

ECE - 2009 7. The time domain behavior of an RL circuit is represented by L + Ri =

0Ω

t=0 0Ω

S

IS

S i(t)

⁄

( + e

i(t) = 0 i(t) = i(t) = 0 i(t) = 0

ECE - 2011 10. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0. The current i(t) at a time t after the switch is closed is i(t)

sint) u(t)

For an initial current of i(0) =

0 2 e 0 2 e 0 e e

, the

0

steady state value of the current is given by

00 0 F

(A) i(t) →

+

(B) i(t) →

8.

(C) i(t) →

(1+B)

(D) i(t) →

(1+B)

The switch in the circuit shown was on position ‘a’ for a long time and is moved to position ‘b’ at time t = 0. The current i(t) for t > 0 is given by 0 kΩ a

b i(t)

100 V

0.2 F

+

0.2e 20e 0.2e 20e

i(t) = exp( 2 0 t) i(t) = exp( 2 0 t) i(t) = 0 exp( 2 0 t) i(t) = exp( 2 0 t)

ECE/EE/IN - 2012 11. In the following figure, C1 and C2 are ideal capacitors. has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is. S

kΩ

t=0

0.3 F

0.5 F

(A) (B) (C) (D)

(A) (B) (C) (D)

u(t)m u(t)m u(t)m u(t)m

i(t)

(A) (B) (C) (D)

ECE - 2010 9. In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i(t) for t 0 is th

Zero a step function an exponentially decaying function an impulse function

th

th


GATE QUESTION BANK

Network Theory

ECE - 2013 Common Data for Questions 12 and 13: Consider the following figure

R k +

1 10V

2

13.

e

⁄

), =

(

e

⁄

), =

( ) (t) = (

e

⁄

), =

msec

( ) (t) = ( 2

e

⁄

), =

msec

( ) (t) =

The current Is in Amps in the voltage source, and voltage Vs is Volts across the current source respectively , are (A) , 20 (C) , 20 (B) , 0 (D) , 20 The current in the 1 resistor in Amps is (A) 2 (C) 10 (B) 3.33 (D) 12

17.

ECE - 2014 14. In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in m ) through the 4 k resistor at t = 0 is _______. k

4k

2

(

( ) (t) =

2A

12.

F

2k

Is

5

Vs

R

2

2

msec msec

A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage , the resistance R, the capacitance C, and the current i(t) is given below: = R i(t) + ∫ i(u)du Which one of the following represents the current i(t)? ( )

k

() 0

15.

16.

+ 0 F

mH

A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are H, F and , respectively, the operating angular frequency (in rad/s) is ________.

0

t

0

t

( ) i(t)

In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t >0?

th

th

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GATE QUESTION BANK

Network Theory

( ) R

R L1

i(t)

+ I

0

R

(A) 1,4 (B) 5,1

t

V

(C) 5,2 (D) 5,4

( )

2.

An ideal capacitor is charged to a voltage Vo and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let

i(t)

=

0

18.

In the circuit shown in the figure, the (t) (in volts )for t → value of is ________ i

0u(t)

2H 2i

, the voltage across the capacitor

at time t > 0 is given by (A) Vo (B) Vocos ( t) (C) Vosin ( t) (D) Voe cos ( t)

t

+

√

+ (t)

EE - 2007 3. In the circuit shown in figure switch is initially CLOSED and Sw is OPEN. The inductor L carries a current of 10 A and the capacitor is charged to 10 V with polarities as indicated. Sw is initially caps at t = 0 and Sw is OPENED at t = 0. The current through C and the voltage across L at t = 0 + is SW2 R2 0Ω

EE - 2006 1. In the circuit shown in the figure, the current source I = 1A, voltage source V = 5V, R = R = R = 1Ω, = = = 1H, = = 1F. The currents (in A) through R3 and the voltage source V respectively will be

R1= 0Ω

SW1

L 10A

(A) 55 A, 4.5 V (B) 5.5 A, 45 V 4.

C

+ _10V

(C) 45 A, 5.5 V (D) 4.5 A, 55 V

The state equation for the current I1 shown in the network shown below in terms of the voltage Vx and the

th

th

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GATE QUESTION BANK

Network Theory

independent source V, is given by 0 2H

1F +

+

1F 0 H

+

1F

3A

02

⁄ s ⁄4 s

(A) (B) (A)

= 1.4 Vx – 3.75I1 + V

(B)

= +1.4 Vx – 3.75I1

(C)

= 1.4 Vx + 3.75I1 + V

(D)

= 1.4 Vx + 3.75I1

V

V

EE - 2008 Statement for Linked Answer Questions 5 and 6 The current i(t) sketched in the figure flows through an initially uncharged 0.3 nF capacitor.

(C) 4 s (D) 9s

EE - 2009 8. In the figure shown, all elements used are ideal. For time t<0, S remained closed and S open. At t = 0, S is opened and S is closed. If the voltage Vc2 across the capacitor at t = 0 is zero, the voltage across the capacitor combination at t=0+ will be S1 S2

3V

C1

1F

C2

2F

6 i(t) mA5 4 3 2 1

(A) 1V (B) 2 V

0 1 9

5.

2

3

4

5 6 t ( s)

7 8

The charge stored in the capacitor at t = 5 µs, will be (A) 8nC (C) 13nC (B) 10nC (D) 16nC

(C) 1.5 V (D) 3 V

EE - 2010 Linked Answer Questions 9 and 10 The L-C circuit shown in the figure has an inductance L = 1mH and a capacitance C = 10µF. L i

6.

7.

The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1s will approximately be (A) 18.8 V (C) 23.5 V (B) 23.5 V (D) 30.6 V

t=0

9.

100 V 100+V

C

The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is: (A) 5 cos( 0 t) A (B) 5 sin( 0 t) A (C) 10cos( 0 t) A (D) 10 sin( 0 t) A

The time constant for the given circuit will be

th

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GATE QUESTION BANK

10.

The L – C circuit of Q.9 is used to commutate a thyristor. Which is initially carrying a current of 5 A as shown in the figure below. The values and initial conditions of L and C are the same as in Q.9. The switch is closed at t=0. If the forward drop is negligible, the time taken for the device to turn off is

EE/IN - 2012 Statement for Linked answer question 14 and 15: In the circuit shown, the three voltmeter reading = 220V, = 122 V, =136 V R R

~

L I C

t=0

100V

(A) (B) 11.

The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0 , the current through the 1µF capacitor is Ω t =0 1µF

(A) 0A (B) 1A

14.

The power factor of the load is (A) 0.45 (C) 0.55 (B) 0.50 (D) 0.60

15.

If RL = 5, the approximate power consumption in the load is (A) 700W (C) 800W (B) 750W (D) 850W

20

(C) 2 s (D) 26 s

5V

oad

100V

5A

2 s 6 s

Network Theory

4Ω

(C) 1.25A (D) 5A

EE - 2014 16. The switch SW shown in the circuit is kept at position ‘ ’ for a long duration t t = 0 the switch is moved to position ‘2’ Assuming , the voltage (t) across the capacitor is R

2

S

EE - 2011 Common Data For Q.No 12 & Q.No 13 An RLC circuit with relevant data is given below =

R

~

0

= √2 = √2

12.

13.

4 4

The power dissipated in the resistor R is (A) 0.5 W (C) √2 (B) 1 W (D) 2 The current (A) 2 (B)

√

R

17.

in the figure above is (C) +

√

(D) +j2A

th

(A)

(t) =

(

e

⁄

)

(B)

(t) =

(

e

⁄

)+

(C)

(t) =

(

(D)

(t) =

(

+

)(

e

⁄

)

)(

e

⁄

)+

The voltage across the capacitor, as shown in the figure, is expressed as (t) = ) sin( t ) + sin ( t The values of and respectively, are

th

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GATE QUESTION BANK H

20 sin 0t

~

(A) 2.0 and 1.98 (B) 2.0 and 4.20

(B) P = 0 and Q = 3/√

F

(t)

Network Theory

(C) P = 5 and Q = 6/√ (D) P = 5 and Q = 3

0 sin t

(C) 2.5 and 3.50 (D) 5.0 and 6.40

IN - 2006 1. In the circuit shown in the following figure, the input voltage vi(t) is constant at 2V for time t s and then it changes to 1 V. The output voltage, v0(t), 2 s after the change will be

IN - 2008 Statement for Linked Answer Question 3 and 4: In the circuit shown below the steadystate is reached with the switch K open subsequently the switch is closed at time t=0.

t=0

H

2

Vi(t)

+ +

F

0

2

3.

1 t

0

At time t=0 , current I is (A) ( ⁄ ) (C) (B) 0A (D)

1µF

+

+

Vi(t)

(A) (B) (C) (D)

M

4.

At time t =0 , is (A) (B)

Vo(t)

5A / s ( 0⁄ )

is (C) 0A / s (D) 5A / s

S

IN - 2010 5. In the dc circuit shown in the adjoining figure, the node voltage V2 at steady state is

– exp ( 2) V – 1 + exp ( 2) V exp ( 2) V 1 – exp ( 2) V

2k

IN - 2007 2. In the circuit shown in the figure, the input signal is ( ) = + cos

+

+

+ (t)

R

⁄

The steady – state output is expressed as (t) = P + Q cos( t ). If R = 2, the values of P and Q are

20 F

k

(A) 0V (B) 1V

(t)

Fig.

0 F

(C) 2V (D) 3V

IN - 2011 6. In the circuit shown below, the switch, initially at position 1 for a long time, is changed to position 2 at t = 0.

(A) P = 0 and Q = 6/√ th

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GATE QUESTION BANK

i

1

1H

0Ω

4A

10.

0Ω

2 t=0

10V

The current i through the inductor for t 0 is (A) e (C) + 2e (B) + e (D) 2 e

Network Theory

In the circuit shown in the figure, initially the capacitor is uncharged. The switch ‘S’ is closed at = 0. Two milliseconds after the switch is closed, the current through the capacitor (in mA) is _____________ S

2k

i (t) t =0 4 F

2k

IN - 2012 7. In the circuit shown below, the current through the inductor is 11. j 0 0

~+

+ ~

0

0 j

(A)

A

(C)

(B)

A

(D) 0 A

R

A

IN - 2014 8. The circuit shown in figure was at steady state for t < 0 with the switch at position ‘ ’ The switch is thrown to position ‘ ’ at time = 0, . The voltage V (volts) across the 0 resistor at time t = 0 is _________________ 2

t=0

analog interface

P Sense harge control

If T = kT

12.

+ 0

In the microprocessor controlled measurement scheme shown in the figure, R is the unknown resistance to be measured, while R and known. is charged from voltage to (by a constant DC voltage source ), once through R in T seconds and then discharged to . It is again charged from voltage to through R in T seconds.

H

R

Sense ischarge

then

(A) R = kR

(

)

(B) R = kR

ln ( )

(C) R = kR (D) R = R

capacitor ‘ ’ is to be connected across the terminals ‘ ’ and ‘ ’ as shown in the figure so that the power factor of the parallel combination becomes unity. The value of the capacitance required in F is___________.

6

9.

4

The average real power in watts delivered to a load impedance = (4 j2) by an ideal current source i (t) = 4 sin ( t + 20 ) is ______________

~

th

60 0 Hz 0 Hz 0 Hz

th

j

th


lnk

GATE QUESTION BANK

Network Theory


Vc(t)

[Ans. D] The transform circuit is shown below

1

(s) +

+

(s)

( )

For the given circuit Where

( )

(S) =

( )

( )

,

Where, R = 4 (S) =

In mathematical form, (t) = tu(t) 2(t T)u(t 2T) + 2(t 2T)u(t 2T)……………

( )

=

,

=

( )

=

v (S) =

2.

4.

( ) ( )

=

⁄ (

⁄ )

(t) = δ(t)

)

(S) = √

(S + ⁄2) + ( ) (t) =

(S) , v (t) = 0 v (t) 0, v (t) =

=(

(S) =

=

5.

⁄

e

√

Sin (

[Ans. B] (S) = R (S) =

(

=

⁄

(

(t) = e 6.

) ⁄

√

( S

(S)

=S

=0 m

( )=0 Time constant of the circuit = R = 4 F 20 k 20 = 40 ms Using direct formula (t) = ( ) ( ) (0)]e (t) = 0 (0 0 )e (t) = 0 e m 3.

nT)

[Ans. D]

[Ans. A] At t = 0 Capacitor is short circuit and at t = apacitor is open circuit So (0 ) =

nT)u(t

F

,

for v (t) = 0 for t

) (t

= tu(t) + 2 ∑(

Where, R = , =4 F R =R = 0 4 0 sec =4 0 sec For t 0u(t) = s = 0 (for dc signal) ( )

t

-1

(s)

(s)

2 T

T

t)

(s))

(S) =

S (S + S + )

⁄ ) (√ ⁄ )

os (

√

t)

[Ans. B] In the circuit shown, R =0 = R (R + R )

√

e

⁄

Sin (

√

= (R + R )

T = ⁄R

[Ans. C] The waveform of voltage Vc(t) is shown below.

= ⁄(R + R ) R ⁄(R + R ) = = = T ⁄(R + R )

di | dt

th

th

th

R


t)

GATE QUESTION BANK

7.

Network Theory 0Ω

[Ans. A] Take L.T of given differential equation with i(0)=

and use the L.T pairs: 1.5 A

sin(t) u(t) → e

0Ω

+ Ri =

( + e

(0 ) = t t = , the status of the circuit is shown in Fig. 3

sin t) u(t)

Take Laplace on both side [s (s)

R

4

, (s + ) +

L

i(0 )

0Ω

s +

sin(t) u(t) → ()

i (0 )

S

] + R (s) =

r ( s + R) (s) = +

s

0Ω

0.5 A

+ (s + ) + +

+ (

0.5 A

0.5 A 1.5 A

)

0

i ( )

S 0Ω

L

i( ) 0Ω

(s) = [

+

+

] (

)

(

)

i( ) = 0 The given circuit is a first order circuit

According to final value theorem, Steady state value of i(t) = i(t) = →

8.

s (s)] =

t

→

R= 0+( 0 = mH =

R

[Ans. B] hen S is in position ‘a’, (0 ) = 00 fter S is moved to b, (for t 0) = 00 =0 R= 0 02 = =0 6 F (t) = 00 e ⁄ (t) i(t) = = 20e

9.

with time constant, =

=

0.75 A

i(t) = =0 10.

1.5 A

0Ω

0Ω 0Ω

15mH

i (0 ) = 0 , i(0 ) = 0 S is closed for 0 ≤ t < At t = 0 , i (0 ) = i (0 ) = 0 The status of the circuit is shown in Fig. 2

)(

+( )( 2 0 2 e

[Ans. A] Q(0 ) =

e ),t e ,t

11.

th

0

) 0

2 m (

)

=

=

For t 0, = 0 = 00 = R = 0 m sec (t) = ( 00 )u(t) 0e d i (t) = dt = 0 0 0 2000 e = e u(t)

i (0 )

i(0 )

+ (F

(0 ) =

0.75 A S

= 0 sec, = 0 sec

i(t) =

[Ans. A] S is open for < ≤0 At t = 0 the status of the circuit is shown in Fig. 1 0.75 A

=

=

General formula:

= 00e u(t)m

0) = 0 + 0 H

0

u(t)

[Ans. D] When the switch in closed at t = 0 Capacitor C1 will discharge and C2 will get charge since both C1 and C2 are ideal and there is no-resistance in the circuit th

th


GATE QUESTION BANK

charging and discharging time constant will be zero. Thus current will exist like an impulse function.

15.

Network Theory

[Ans. *] Range 0.45 to 0.47 A series LCR circuit operating frequency is such that current leads the supply voltage so The current at resonance =

12.

[Ans. D]

R=

2

v

= H

= amp

= F

The magnitude of current is

the value at

resonance Ω

0 +

10V

+ v

0

=

= 0

(

+

=0

0+2=

=

)

)

]

as

√

+√

=0 √ +√ +4 = 2 = 0 4 rad/sec (omitting the negative value)

=

= 0

[Ans. *] Range 1.2 to 1.3 k

4k

k

+

16.

+

k

0 urrent after long time = = m 0k At t = 0 the voltage across the capacitor is 10 Where, = = 0 = Now at t= 0

i i(t)

+

i F

2k

= 2000i(t) dv di i =c = 0 002 dt dt di i = [0 002 + i(t)] amp dt di = 2 + 000i(t) dt + = di 2 + 000i(t) + 2000i(t) = dt s di 2 + 000i(t) + 2000i(t) = dt s 000 2 [s (s) + (s)] = 2 s

4k

The current is m = 4

[Ans. A]

mH

0 F

+

√ +(

) =

13A

[Ans. C]

0

=

x )

2=

+

2

√R + (x

2=√ +(

current through 14.

=

Ω

Across AB voltage drop is 10V = 13A, = 20V = 13.

0 2 =

2A

_

20

2Ω

2 m

th

th

th


GATE QUESTION BANK

000 ]= 2 s

2 (s) [s + (s) =

2s(s +

= [ 2 s =

[

600 s

= [

s

i(t) =

17.

By Nodal analysis at node A 0+

00

s+ s+ e

+i =0

0+

00)

s+

Network Theory

]

00

= 0

00

00

i =

(t) = i =

] amp EE 1.

]m

= *

e

+ where is in m sec

= [

e

], =

=

=

2

[Ans. D] At steady state, inductor acts as short circuit & capacitor acts open circuit 2Ω

]

2

+i =0

msec

Ω

_+ 5V

[Ans. A] = Ri(t) + ∫ i(u)du =R + =

S =

Current through R = 5/1 =5A Current delivered by 5V source = 5 1 = 4A

aplace transform ] 2.

is a D.C voltage source, so

=

where

depends on

S k = = s(RS + ) R (s + i(t) =

[Ans. B] The relevant circuit is shown in fig. +

+

(t)

L

C

)

k e R

Fig.

=

()

It is a standard LC circuit. With v (t) = cos( t) or sin( t + 0 ) 3. 18.

[Ans. D] Equivalent circuit at t = 0 is,

[Ans. *] Range 31.24 to 31.26

V

i + 0u(t)

2H 2i

0Ω

+

L

I 0Ω 10A

+

10V

(t)

By nodal analysis, 0+ Since we have to evaluate (t) in steady state the inductor will behave as a short circuit and hence = i

=

th

th

(

)

=0 =

th

2 (

0 = 00 )

= 4


GATE QUESTION BANK

4.

[Ans. A]

8.

V-3( +

)

0

0

[Ans. A] The status of the circuit at t = 0 shown in fig. 1.

= 0 …… (1)

= 0.2

0

+02

Network Theory

A

= 0……… (2) + 3V

Eliminating I2 for eq(1) and (2) we get = 5.

4

3V

+

Fig. 1 The status of the circuit at t = 0 shown in Fig. 2. S S A

2 A1

s

s

B

Charged stored in the capacitor = area under i – t curve Q= + = (2 2

0 )

(4 0 0

= [4 +

6 2

]

0

0 ) + (4 + 2) 2 ( 2) =

= 9.

n

(0 ) 00 = s s 1/s C

I(s )

(s) =

cos

06 7.

=

sL

0

√

=

=

(s + 00

)

( s +

)

[Ans. C] = F;R

= 6Ω

T=R

Fig. 2

[Ans. D] Initial current through the inductor is zero and capacitor voltage is charged upto to voltage (0 ) = 00 As current through inductor and voltage across cannot change abrupty. So, after closing the switch i (0 ) = i (0 ) = 0 And (0 ) = (0 ) = 00 The circuit is s – domain

[Ans. D] apacitor charged upto s, so total charge stored in capacitor = Q = 13nC Voltage across the capacitor before connecting to indictor Q 0 = = =4 0 0 Voltage across the capacitor at time t (t)at t = s (t) = cos t] = 2.357 rad = (t) =4

is

+ V

+ 3V

3V

A2

2 s

t=

2F

1F B

[Ans. C] i(t) mA 4

6.

is

⁄√

= 00√ (

= 4sec

s + ( ⁄√ th

th

th

) )


GATE QUESTION BANK

Taking inverse Laplace transform (s)] i(t) =

= √2 13.

= 00√ sin

√

t

= 00

√

sin (

t)

√

i(t) = 10sin (104t)A 10.

14.

[Ans. A] i iT

Network Theory

VC(t)

√2

[Ans. D] Using KCL, + + = = √2

=

=0 √2

=2 0 =+ 2 [Ans. A] Pharor diagram = 220 = 122 V = 136V ⃗⃗⃗ = ⃗⃗⃗ + ⃗⃗⃗

20

100V

5A

When the switch is closed, I flows through thyristor. Load current = = Net current through thyrostor i = i i = sin 0 Let at t=T, circuit get turned off and current i becomes zero i = 0 sin 0 T = 0 0 sin T = sin 0 T = 0 0 T = 0 or 0 2 rad T 2 s 11.

[Ans. B] (0 ) = 4 i (0 ) =

12.

(

(0 ) = 4 )

𝛟 I

By parallelogram law of addition of vectors = + +2 cos by using options, cos = 0.45 15.

P 16.

= A

[Ans. B] Power supplied by the source = cos Where = angle between = 4 inductor and capacitor do not consume power. Therefore, power dissipated in R = Power supplied by the source P = cos =

√2

cos

[Ans. B] :R = = cos = 136 x 0.45 [From Q 4] = 61.2V =

= 749W = 750W

[Ans. D] At t = 0 Voltage across capacitor = tt 0 +i R+i R+ =0 d + 2c R+ =0 dt Taking Laplace s s

+ (2Rc) s + 2Rcs

(0)] + 2R

+

( + 2Rcs) = 2R

th

=0 s

4

th

=0

th


GATE QUESTION BANK

2R + 2Rcs S( + 2Rcs) 2R 2Rc (s) = + + 2Rcs + 2Rcs s

1µF

(s) =

(t) +

(e) =

* (t) = 17.

(

e

+

)(

+

Network Theory

+

+

Vi(t)

M

Vo(t)

e

e

)+

Input is 2V for t 1s. Time constant ( ) =

+

= =

0

0

So (0 ) = 2 Laplace Equivalent circuit after the change 2/s 1/sc

[Ans. A] Applying superposition theorem Resolving 0 sin t source

+

(t) =

sin ( 0t + tan √

( )

0)

( )

Removing 20 sin 0t voltage source

1/s

+

R

I(s)

H

F

s

(s) =

R+

2 s

s

=

(s) =

=

+

=

Rsc + c (s) = = ce +s Now voltage across Resister at t = 2s (t) = (t) R ce R ( ) = = Rc e e ] ( ) =

+ j ( )

= 0 sin( t)

+

= 0 sin( t) 0

=

√ +2 (t) =

+ j

sin t

tan ( )] 2.

20 √ 00 +

( 0 tan +

IN 1.

[Ans. A] The circuit is shown in Fig.

sin( 0t

0 √ +2 tan )

(

0

)) +

sin( t

+ (t)

(t)

R

Fig.

[Ans. A]

(t) = + cos( t) Due to 5V d.c alone , C is open and no current flows through R, = 0 and =0 ue to cos ( t) alone, with

Vi(t)

2 1 0

t

THE GATE ACADEMY tPVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 182 th

th

th

GATE QUESTION BANK

H( ) = For

=

R

=

R+

j R +j R

iL (t) andv (t) cannot change instantaneously. i (0 ) = (0 ) = (0 ) = 0 v (0 ) = v (0 ) = 0 At t =0 , the state of the circuit is shown in Fig. 4

R=2 R

H( ) =

2

=

R √ + √ H( ) = 0 tan (2) v (t) =

√

cos

t+ 0

I1(0 )

C

(

After a long time, at t = 0 L behaves as short circuit and C behaves as open circuit. The relevant circuit is shown in Fig.2. IL(0 ) L

2

2k

At t = 0 , I1(0 )=0, iL(0 )= I2(0 )=0 and v (0 ) = 0 For t > 0, the switch, K is closed and the relevant circuit is shown in Fig. 3 1

V2

R2

20 F

Fig. 1 2k

C1

V1

V2

R1

I2

I

2

1k

9V

I1 5V

0 F

V1

1k

9V

Fig. 2(at t = 0 )

K

A

R1

10 V

C

(0 )

= (0 ) =

[Ans. B] The given circuit is shown in Fig. 1 t steady state i e , as t → , capacitor behaves as open circuit. The circuit at steady state is shown in Fig.2

I2(0 ) I1(0 )

)

5.

10 V

I=0

10 V

[Ans. B] From Fig. 3: Write the Outer loop equation: d (t) (t) ]+ + 0= dt At t = 0 d (0 ) = (0 ) = + dt = /sec

Fig. 1(for t <0)

1

0 = v (0 )

4.

1H

1F

i (0 ) = 0

Fig. 4

I1 2

L

2

5V

I2

I=0

I(0 ) I2(0 ) = 0

5V

[Ans. A] For t < 0, the circuit is shown in Fig. 1 1

1

K

tan(2)]

According to Superposition Theorem v (t) = v (t) + v (t) v (t) = P + Q cos( t ) 6 P = 0, Q = , = 0 + tan (2) √ 3.

Network Theory

C2

1H Fig. 2

1 F Fig. 3(t >0)

10 V

Apply voltage division across R1 = 2k and R2 = 1k

th

th

th


GATE QUESTION BANK

7.

Node voltage, V1= =

[Ans. C] (

=

0 0

From the given circuit, = j = ( )

10 V ±

=0

= 8.

i

+

j

= 10

0 + 0

[Ans. D] For t < 0, the status of the circuit is shown in Fig. 1 inductor behaves as short circuit after a long time. i(0 ) =

) j

V1 is divided between C1 = 0 F and C2 = 20 F Apply, again voltage division across C1 and C2 0 = = = + 0 6.

Network Theory

[Ans. tt=0

]

1H Fig. 1

for t 0, the status of the circuit is shown in Fig. 2 Current through the inductor cannot change instantaneously. i(0 ) = i(0 ) = initial value (I.V) After a long time inductor behaves as short circuit. i( ) = 2 = final value (F. V) Time constant =

=

6

6 = 2 At t= 0 T =

, = 20

+ 0

+

i

10

10

(0 )

=

2

10 4A

(inductor in steady state)

+ 0

6

1H

=

0

=

0 ]

Fig. 2

i(t) =

+ (F

for t 0 = + (2 =2 e

)( )(

e

e

( )

9.

[Ans. 32] Load impedance = = (4 Ideal current source i(t) = 4 sin ( t + 20 ) Average real power = i

)

)

=( ) √

Power =

th

th

6 2

j2)

4

4= 2

th


GATE QUESTION BANK

10.

[Ans. *]Range 1.5 to 1.6 at t = 0 , switch is open so (0 ) = at (t = )so capacitor is in steady state

11.

[Ans. C] Time constant of RC Circuit is = R In case I Time constant T = R In case II Time constant (T ) = R Given T = k T So R =k R R =k R ]

12.

[Ans. *] Range 186 to 188

2k

2k

( ) = 2k

4k

Network Theory

( )

=2

We know that (t) = (0 ) v ( )]e = 0 2 ]e +2

4

+

( ) 60

= (2k

)

4

2k = k 4 = 0 4 0 =4 0 s 000 = =2 0 4 (t) = 2 e +2 (t) = 2 2 e we know that dv (t) i (t) = dt d i (t) = c 2 2 e ] dt = 2 c( 2 0)e =4 F i (t) = +2 4 0 2 0e i (t) = 2 0 e (t) ] i = m

0 Hz

0 Hz H 0

j

0 Hz

This is a tank circuit configuration of a Parallel R-L-C circuit. Resonant frequency of tank circuit √

f =

=4

f =

=

(

+(

)

)

=

R=4 , f= 0H +

( )

=

=

= =

2

0

=

th

F

th

th


GATE QUESTION BANK

Network Theory

Sinusoidal Steady State Analysis ECE - 2007 1. The RC circuit shown in the figure is R

C +

+

ECE - 2008 4. The Thevenin equivalent impedance ZTh between the nodes P and Q in the following circuit is 1

1

C

R

1

(A) (B) (C) (D)

1

1

1

a low-pass filter a high-pass filter a band-pass filter a band-reject filter

(A) 1

(C) 2 + s +

(B) 1 + s+ 2.

3.

An independent voltage source in series with an independence ZS =RS + jXS delivers a maximum average power to a load impedance ZL when (A) j (C) j (B) (D) j In the AC network shown in the figure, the phasor voltage VAB (in Volts) is

(D)

ECE - 2009 5. An AC source of RMS voltage 20V with internal impedance Zs (1 2j) Ω feeds a load of impedance ZL (7 4j) Ω in the figure below. The reactive power consumed by the load is (1

2j)Ω

A

2 ∠

5∠3 j3

B

(A) 0 (B) 5∠3

(7

5

5

(C) 12.5∠3 (D) 17∠3

(A) 8VAR (B) 16 VAR

j3

4j)Ω

(C) 28 VAR (D) 32 VAR

ECE - 2010 6. The current I in the circuit shown is 2 m

~

2 ∠

1 1

(C) 0 A (D) 20 A

(A) –j1 A (B) j1 A

th

th

5

rad s

th


GATE QUESTION BANK

7.

For the parallel RLC circuit, which one of the following statements is NOT correct? (A) The bandwidth of the circuit decreases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value

ECE/EE - 2013 8. A source v (t) cos 1 t has an internal impedance of (4 + j3) . If a purely resistive load connected to this source has to extract the maximum power out of the source , its value in  should be (A) 3 (C) 5 (B) 4 (D) 7 ECE/EE/IN - 2013 9. In the circuit shown below, if the source voltage Vs= 10053.130 V then the Thevenin’s equivalent voltage in volts as seen by the load resistance RL is 3

j4

(A) 100900 (B) 80000 10.

j6

j4 

Network Theory

ECE - 2014 11. A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is (A) (18 j 1.5) k (B) (18 j 1.5) k (C) (2 j 1.5) k (D) (2 j 1.5) k 12.

A periodic variable x is shown in the figure as a function of time. The rootmean-square (rms) value of x is ______. x 1

t T 2

13.

T 2

The circuit shown in the figure, the value of capacitor C (in mF) needed to have critically damped response i(t) is ________. 4

4

5 i(t)

1

RL= 10

(C) 800900 (D) 100600

Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency is (A) q q (B) (1⁄q ) (1 q ) ) ⁄( (C) (q q ) ) ⁄( (D) (q q )

14.

In the magnetically coupled circuit shown in the figure, 56 % of the total flux emanating from one coil links the other coil. The value of the mutual inductance (in H) is ______ . 1 6 cos (4t 3 )

15.

th

(1⁄16)

~

4

5

The steady state output of the circuit shown in the figure is given by

th

th


GATE QUESTION BANK

Network Theory

V1 at a frequency which causes resonance with a current of I.

( )). If the y(t) = A( )sin( t amplitude | ( )| 0.25, then the frequency is

I

y(t) sin t

~ ( ) ( )

1

( )

√3 2

( )

1

The phasor diagram which is applicable to this circuit is

2

(A)

√3

I

EE - 2006 1. In the figure the current source is 1 ∠0 A, R = 1Ω, the impedances are Zc = j Ω, and ZL = 2jΩ. The Thevenin equivalent looking into the circuit across X-Y is

(B)

I

x

(C)

y

(A) (B) (C) (D)

√2∠0 V, (1 + 2j) Ω 2 ∠450 V, (1 – 2j) Ω 2 ∠450 V, (1 + j) Ω √2∠450 V, (1 + j) Ω

I

(D)

2.

3.

An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak to peak square wave source of 150 Hz. The power in kW dissipated by the heater will be (A) 3.478 (C) 1.540 (B) 1.739 (D) 0.870 The circuit shown in the figure is energized by a sinusoidal voltage source

I

EE - 2007 4. In the figure transformer T1 has two secondaries, all three windings having the same number of turns and with polarities as indicated. One secondary is shorted by a 1 Ω resistor , and the other by a 15m th

th

th


GATE QUESTION BANK

capacitor. The switch SW is opened (t =0) when the capacitor is charged to 5 V with left plate as positive At (t =0+) the voltage VP and Current IR are

A

T1

L

R B

+ +

Vp

Im

(A) 25 V, 0.0A (B) Very large voltage, very large current (C) 5.0 V, 0.5 A (D) 5.0 V, 0.5 A 5.

C

(A)

C

25V

R

~

IR SW

Network Theory

Re

(B)

In the figure given below all phasors are with reference to the potential at point “O”. The locus of voltage phasor YX as R is varied from zero to infinity is shown by

Im

Re

~

∠

R

(C) C

XY

~

∠

Im

O 2V

(A) 0

(C)

2V 0

VYX

VYX Locus of VYX

Re

Locus of VYX

(D) Im

(B)

Locus of VYX

(D)

Locus of VYX

VYX 0

VYX 2V

0

2V Re

6.

The R-L-C series circuit shown is supplied from a variable frequency voltage source. The admittance locus of the network at terminals AB for increasing frequency is

EE - 2008 7. The Thevenin's equivalent of a circuit operating at = 5rad/s, has = 3.71∠ 15.90 V and ZO =2.38 – j 0.667Ω. At this frequency, the minimal realization of the Thevenin's impedance will have a th

th

th


GATE QUESTION BANK

(A) resistor and a capacitor and an inductor (B) resistor and a capacitor (C) resistor and an inductor (D) capacitor and an inductor

Network Theory

EE - 2011 10. The r.m.s value of the current i(t) in the circuit shown below is 1

1 1

8.

The resonant frequency for the given circuit will be

i(t)

1

0.1H

~ (1. sin t)

(C) 1 (D) √2

(A) (B) 1

1

√

Common Data Q. 11 and Q. 12 The input voltage given to a converter is (A) 1 rad /s (B) 2 rad /s

(C) 3 rad /s (D) 4 rad /s

EE - 2010 9. If the electrical circuit of figure (b) is an equivalent of the coupled tank system of figure (a), then

v 1 √2 sin(1 t) The input drawn by the converter is i

11.

h

(a) oupled tank B A

3

) t

2√2 sin(5 t The input power of the converter is (A) 0.31 (C) 0.5 (B) 0.44 (D) 0.71

4

) 6)

The active power drawn by the convert is (A) 181 W (C) 707 W (B) 500 W (D) 887 W

and C, D

EE - 2012 13. A two–phase load draws the following phase currents: ), i (t) sin( t i (t) cos( t ). These currents are balanced if 1 is equal to ) (A) (C) ( 2 ) (B) (D) ( 2

and B, D

14.

D C

(b) Electrical equivalent

(A) A, B are resistances capacitances (B) A, C are resistances capacitances (C) A, B are capacitances resistances (D) A, C are capacitances resistances

t

5√2 sin(3

12.

h

(1 √2 sin (1

and C, D and B, D

th

The average power delivered to an impedance (4 j3)  by a current 5 cos (100t+100) A is (A) 44.2 W (C) 62.5 W (B) 50 W (D) 125 W

th

th


GATE QUESTION BANK

EE - 2014 15. A combination of 1 µF capacitor with an initial voltage ( ) 2 in series with a 100 resistor is connected to a 20 mA ideal dc current source by operating both switches at t = 0s as shown. Which of the following graphs shown in the options approximates the voltage across the current source over the next few seconds?

Network Theory

16.

A non-ideal voltage source has an internal impedance of . If a purely resistive load is to be chosen that maximizes the power transferred to the load, its value must be (A) 0 (B) real part of (C) magnitude of (D) complex conjugate of

17.

A series RLC circuit is observed at two frequencies. At = 1 krad/s, we note that source voltage 1 ∠ results in a current . 3∠31 A. At = 2krad/s, the source voltage 1 ∠ results in a current 2∠ A. The closest values for R,L,C out of the following options are (A) 5 25 m 1 (B) 5 1 m 25 (C) 5 5 m 5 (D) 5 5m 5

t t

( )

t 2

IN - 2006 1. Consider the AC bridge shown below. If

( )

RC = 1 and

V ΔC <0.01, then ratio 0 is C Vs

approximately equal to

t

R

2

( )

R

C

t

~

2

( )

(C)

(A) √1

(D)

(B)

t 2

IN - 2007 2. In the circuit shown in the following figure, the current through the 1Ω resistor is th

th

th


GATE QUESTION BANK

(B)

j j

L=1H C = 0.25F

1Ω

V = 1V

3.

j

I = (5cos2t)A

(A) (1 + 5 cos 2t) A (B) (5 + cos 2t) A

(C) (1 5 cos 2t) A (D) 6A

Consider the AC bridge shown in the figure below, with R, L and C having positive finite values Then

if

(B)

if

(C)

if

(D)

j ( j ( j

) )

(D)

j ( j ( j (

) ) )

1 √2 sin (1

t)

1Ω

1 √2 cos (3

t)

1 mH

(A) 50 W (B) 1050 W

(C) 5000 W (D) 10100 W

√

cannot be made zero

6.

For the circuit shown below the voltage across the capacitor is

Consider the coupled circuit shown below.

1 Ω

~

j1

(10+j0)V

j1

Ω

Ω

ig a

At angular frequency , this circuit can be represented by the equivalent T – network shown below.

(A) (10 + j 0)V (C) (0 + j100)V (B) (100+j0)V (D) (0 – j100)V For the circuit shown below the steadystate current I is

7.

cos(1000t)V

I

ig b

Indicate the correct set of expressions for the impedances of the T – network (A) j ( ) j ( ) j

1Ω

V(t)= 5√2

4.

(C)

IN - 2008 5. In the circuit below the average power consumed by the 1 resistor is

sin t

(A)

Network Theory

th

1mH

th

th

1000 F


GATE QUESTION BANK

Network Theory

(A) 0A (B) 5√2 cos (1000t)A (C) 5√2 cos (1

t

)

(D) 5√2A IN - 2010 Statement for Linked Answer Questions: 8 &9 A coil having an inductance (L) of 10mH and resistance R is connected in series with an ideal 100 F capacitor (C). When excited by a voltage source of value 10√2 cos (1000t)V, the series RLC circuit draws 20W of power. 8.

The value of the coil resistance R is (A) 1 (C) 4 (B) 2 (D) 5

9.

The Q factor of the coil at an angular frequency of 1000rad/s is (A) 1 (C) 4 (B) 2 (D) 5

IN - 2011 Common Data for Question 10 & 11 Consider the circuit shown below:

2 sin(5t)

2

cos(5t)

.1 i(t)

10.

The current i(t) through the capacitor is (A) sin (5t) A (C) sin (5t – 45) A (B) cos (5t) A (D) 1 A

11.

The average total power delivered by the sources is (A) 0 W (C) 2 W (B) 0.5 W (D) 4 W

th

th

th


GATE QUESTION BANK

Network Theory


6.

[Ans. A]

[Ans. C]

j

( || 1⁄ )

( ) ( )

1⁄

(

2 ∠

(

J ,

(

)

(

)

3

1)

(

)

(J ) (J )

, ,

J 1⁄ 3 3J Filter is band-pass filter. 1⁄

t

2.

power

j

1

*

, (J )

[Ans. D] For maximum J

1

1

(L = 20 mH, C = 50 ) Nodal analysis at node A 2 j 1 1 j

(J ) s s

~

1

( ) ( ) ut

( || 1⁄ ))

j1 j [ 1 2 j1

transfer,

2 2

j1

1 j ] 2

5

1

+

j1

1 3.

[Ans. D] current 53 5 3

4.

j1 impedance ((5 3J)||(5 34 173 1

3J))

[Ans. A] Replace 10V by short circuit and 1A by open circuit ( 1). (1 1⁄ ) 1 1⁄ 2

7.

[Ans. D] This is standard concept of parallel resonant circuit

8.

[Ans. C] |

9.

[Ans. B] The RMS current in the load is given by ⃗ ∠ tan ( ) 2∠

tan

5Ω

3

j4

∠53 13 5∠53 13 j4 2 ∠ 4∠ 2 ∠ 8 9 8 9 j4

( )

2 , reactive power 4 4 16 Also note that the active power consumed by the load 4 7 28

3

[Ans. C] 1 j4

5.

√4

|

10.

1

[Ans. C]

fter series connection th

th

th


GATE QUESTION BANK

(

1 2

) 1

11.

[Ans. B] Total ower (s) 1 ∠ cos .8 .8 18 1.5j

1 ∠ cos .8

4 c

.8

[Ans. *] Range 0.39 to 0.42 x

1 2

√

1 4 √ 4

c 12.

Network Theory

14.

(T⁄2 , 1)

1

1 2

4 4 4

1

[Ans. *] Range 2.49 to 2.52 oupling coefficient given is e know,

t T 2

iven

T 2

Equation of line 1 (t T ⁄2 2t T 1 √ *∫ T

15.

4 3

5

[Ans. B]

2t ( ) dt T

y(t) sin t

+ ( ) sin[ t ( )] y(t) | ( )| .25 By nodal analysis x 1∠ x x ⁄ 1 cj 2 cj 1 cj 1∠ x[ cj ] 2 2 3 cj 1∠ x[ ] x 2 x 1 y 2 2 3 cj 1 1 | ( )| 4 √4 9 9 12 2

1 8

√6 .4 8 [Ans. *] Range 9.99 to 10.01 4

4

5

x

1

13.

4

)

1 4t √ * + T 3T √

.56 √

√

.56 √4 2.5 4

1 √ ∫ x dt T

1 m

1

i(t)

For critically damped 1 1 2

2

2 3 cj

√3 c

th

th

th


GATE QUESTION BANK

EE 1.

V

[Ans. D] To calculate Thevenin’s impedance, current – source is open – circuited x R=1

200 V 2

T/2

Zth

ZL=2j

T

ZC= j

z

z z 1 2j j = 1 j Open – circuit voltage at terminals X – Y z (1 j) 1

1 * ∫ T

1 15

dt+

1 ∫2 dt ∫( 2 T [ { 2 2 1 k 23 1.739 k

√245 volts [Ans. B] Assuming resistance of the heater = R (i) When heater connected to 230 V. 50 Hz source, energy consumed by the heater = 2.3 units or 2.3 kWh in 1 hour Power consumed by the heater

1 f

Vrms value of the input voltage

y

2.

Network Theory

3.

) dt }]

[Ans. A]

2.3k h 1 hour 2.3 k rms value of the input voltage 23

j(

)

At resonance,

2.3

1

23

so Therefore, input impedance is purely resistive, is minimum and the input voltage and output current are in phase

23

(ii) When heater connected to 400 V (peak to peak ) square wave source of 150 Hz

(

)

[

j(

)]

phase

with

but

Therefore,

is

in

Voltage across the capacitor

th

th

th


GATE QUESTION BANK

1

5.

Network Theory

[Ans. A]

j l

∠

9

R

∠

So, lags the current by 9 . The phasor diagram on the basis of above analysis

x

y C

∠

O

Let capacitive reactance ∠

4.

[Ans. D] R T1

∠

using

IR

,.

s1 Vs s2

(

P1 25V

(

VP

) )

C P2

(

T1 T2

)

(

VT

)

Method -1

All the three windings has same number of turns, so magnitude of induced emf’s in all the three windings will be same i.e. | | | | | | Polarity of the windings is decided on the basis of dot – convention. As capacitor is charged to 5 V with left plate as positive. So, T1 is positive wrt T2 5 As T2 has negative polarity. So, P1 has negative polarity Therefore, VP = 5 Similarly, S1 has negative polarity So, 5 5 .5 1

*

+

*

+

*

+

*

+

when

when Method-2

∠18 ∠ (18

√

∠

√

∠

( 2 tan

( (

)

)

)

( ))

Magnitude of So, option (c) and (d) can not be correct, as magnitude is 2V in these two options.

th

Angle of

1 8

When ∠ 18 18 when

2tan ( ) 2 9 36

th

th

2 tan

( )


GATE QUESTION BANK

Network Theory

∠ 18 2 tan ( ) 18 On the basic of above analysis, the locus of is drawn below: 2V

O when

∠

Ref when R=0 √

6.

[Ans. B] Admittance of the series connected RLC (

7.

[Ans. B] Thevenin’s mpedance 2.38 j .667  as real part is not zero , so has resistor m[ ] j .667 Case –I Z0 has capacitor (as Im[Z0]is negative ) Case – II Z0 has both capacitor and inductor, but inductive reactance  capacitive reactances at = 5 rad / sec For minimal realization case – (i) is considered Therefore, Z0 will have a resistor and a capacitor.

8.

[Ans. C] Input impedance 1 z j ‖ j

)

(

)

(

)

*By rationlization} Separating, real and imaginary part of admittance. e[ ] (

)

For any value of positive. When At

, the real part is always or (resonance)

√

e[ ]

(maximum value) (

( )

)

( (

) (

At

√

)

)

z

(resonance)

Imaginary part of zero m( ) For

j

1

z

Therefore Im [Y] > 0 For

j 0.1H

1F

z

1

j .1 j .1

Therefore, [ ] On the basis of above analysis, the admittance locus is

1 j ( .1 ) 1 1 At resonance, imaginary part must be zero. .1

th

1 th

th


GATE QUESTION BANK

.1

1

11.

1 1

1 9 3 rad/sec

9.

l

C=

10.

[Ans. B] rms value of input voltage 1 √2 1 √2 rms value of current 1 √2 5√2 2√2 √( ) ( ) ( ) √2 √2 √2 11.358 Let input power factor cos l cos =active power drawn by the converter 1 11.358 cos 5 cos .44

[Ans. D] In such system, volumetric flow rate C is analogous to current and pressure is analogous to voltage. The hydraulic capacitance due to storage in gravity field is defined as Where A = Area of the tank ρ ensity of the fluid g = Acceleration due to gravity The hydraulic capacitance is represented by A & C. Liquid trying to flow out of a container, can meet with resistance in several ways. If the outlet is a pipe, the friction between the liquid and the pipe walls produces resistance to flow. Such resistance is represented by B & D.

Network Theory

12.

[Ans. B] 1 i

√2 sin(1

t)

1 √2 sin (1

t

3

)

5√2 sin (3

) 4 2√2sin(5 t 4) Fundamental component of input voltage ( ) 1 √2 sin(1 t) 1

( )

,

(i )

,

t

√2

1 √2 Fundamental component of current (i ) 1 √2 sin(1 t 3)

[Ans. B] 1 sint sin t 1 and 1 rad sec Impedance of the branch containing inductor & capacitor j( ) 1 j( ) 1 1 j (1 1 ) 1 1 So, this branch is short – circuit and the whole current flow through it 1. sin t i(t) 1. sint 1 rms values of the current 1

1 √2

√2 Phase difference components

1 between

these

two

, cos cos .5 3 3 Active power due to fundamental components ( ), (i ) , cos 1 1 .5 5 Since, 3rd & 5th harmonics are absent in input voltage, there is no active power due to the these components. Hence, active power drawn by the converter =Active power due to fundamental components = 500 W

√2

th

th

th


GATE QUESTION BANK

13.

[Ans. D] i sin( t i cos( t

17.

[Ans. B]

) )

j(

sin (

t ) 2 As these two currents are balanced. i i sin

sin

2. sin ( )

sin( t 2 sin ( t

14.

4

2

1k rad s 1 ∠ . 3∠31 1 ∠ . 3∠31 j( )

)

j (1

) t

)

1

2

1

1

4

1 1

2 (2

α)

5

IN 1.

4

1 and 25

(

)

(

) )

1

d dt

1 ∫ dt (

k

At t = 0, k 2 t t

2

16.

)

1

[

1 1

1 2

1 1 2 1

1

t

1 2

)

(

2

2

.

[Ans. D]

(

t

5 ∠

∠

c

)

1

[Ans. C] tt

∫d

∠

)

On solving equation 5 , 1 m ,

[Ans. B] Z = 4 j3 = RL jXc; RL=4; 5cos(1 t 1 ) mcos(cot

1

.

5 2

5

1 ∠ . 3∠31

) 1 c 2k rad s

j (2

) )

)

t

) . cos ( t

2

P= 15.

2

sin (

1 cos ( 2 2 1 ( 2 2

Network Theory

1

2

1

2

1

t

[Ans. C] Magnitude of | | (using transfer theorem)

1 1 2 maximum

1 2

power

th

th

1

]

1 1 1 2 1 2 1

1

1

1

1 2

1 2

2 1 2

( ) 1

2

th


GATE QUESTION BANK

2

2

2 (2

)

i 3.

i

Network Theory

(1

5 cos 2t)

[Ans. D] The A.C bridge is shown in Fig

. 1, it can neglected in comparison to 2 [Ans. A] The circuit is shown in the figure 1

Let

1

.25

i

(5 cos 2t)

1

1

Fig. 1

Superposition theorem can be used to find i Due to 1 V d.c source alone, the circuit is shown in Fig. 2, where the current source is open circuited (I = 0). Also, in steady state, L behave as short circuit and C behave as open circuit

1

,

1

,

,

j√

1

With R, L and C having positive finite value, becomes imaginary which is not possible. cannot be made zero 4.

1

,

j , of the vridge is balanced if product of the impedances of opposite arms is same i.e., j or j

[Ans. A] The coupled circuit is shown in Fig. 1

1 i Fig. 2

From figure 2, i 1 Due to AC source (5 cos 2t) alone, the circuit is shown in figure 3, where the voltage source is short circuited (V = 0)

ig 1

Write the mesh equations: j j (1) j j (2) The equivalent T network is shown in Fig .2

1

1 .25

i

(5 cos 2t)

2.

Fig. 3

For As

2, 2 1

2

.25

.5,

2 ig 2

, the parallel RLC circuit is

Write the mesh equations: ( )

under resonance i 5 cos 2t th

th

th


GATE QUESTION BANK

(

or

)

. (3)

For C = 1000 1 1 l

( ) ( ) or . (4) omparig (1)with (3)and (2)with (4, ) j , j , j ( ) j , j ( ) 5.

[Ans. D] Let v (t) v (t)

1 √2 in (1 1 √2 cos(7

RMS value of v (t)

j(

t) )

√

√ √

1

MSV of v (t) 1 1 Similarly M.S.V of v (t) 1 Let v(t) = v (t) v (t) . . of v(t) 1 1 1 1 1 1 ,1 Power is consumed or dissipated only in 1 . if it is , . . of v(t) 1 ,1

8.

2

=

∠

1∠

)

= 10

circuit it under

1

( )

Q= √

A

1

5Ω

[Ans. B]

[Ans. D] Circuit is under resonance I=

l

[Ans. D] = 1000 1 resonance 1 1 1 1

9. 6.

1

The circuit is open for the excitation frequency 1 r s steady state current, It may also be noted that in steady state, as t v(t) = 0 because of the exponential,

t) and

(

Network Theory

10.

100JV

√

2

[Ans. A] The given circuit is shown in fig. 1 2

cos(1

1

5√2e

)

( ) 2 sin(5t)

[Ans. A] The circuit is shown in the figure

1m

Given:

For the given input, v(t) 5 √2e cos(1 1 r s Z=1+ For L = l mH 1 1 1

.1 i(t)

i( ) cos(5t)

Fig. 1 1

R=2 ,

5 rad sec .1 ,

2

Taking cos (5t) = Re[1e e ] 1 cos (5t) e[1 e e ] ⃗⃗⃗ hasor, 1e for (t) 2 sin(5t) 2 cos(5t hasor, ⃗⃗⃗⃗ 2 e

v(t)

7.

t)

9

)

The phasor equivalent circuit is shown in fig – 2

th

th

th


GATE QUESTION BANK

Network Theory

9 j2

2

⃗⃗⃗

⃗⃗⃗⃗ 2∠

j2 ⃗ Fig. 2

The circuit is further simplified to the circuit in Fig. 3 and Fig. 4 j1

2

1

j2 ⃗

Fig. 3

(1

j)

2

j2 ⃗

Fig. 4 From Fig 4 (1 j)2 (1 j) √2∠45 ⃗ 2 j2 1 j √2∠ 45 1∠9 1∠ 9 j ⃗ i(t) e[ e ] e[1e e ]

e[e ( sin(5t) 11.

)

]

cos(5t

[Ans. C] Phasor current through 2

1 1 1

9

)

j j

j

i (t) 1 cos(5 t) Power is dissipated in the resistance, 2 . Assuming that the source values are given in RMS value Power delivered by the 2 sources = Power dissipated in the equivalent 2 of Fig. 4 (1) 2 2 Note: Assuming Peak value for the sources, Power delivered by the 2 sources =( ) √

2

1

th

th

th


GATE QUESTION BANK

Network Theory

Laplace Transform ECE - 2006 1. The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by (A) RL network only (B) RC network only (C) LC network only (D) RC as well as RL networks 2.

A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is I(s)

ECE - 2009 4. If the transfer function of the following network is

= R

+

+ RL

Vi

V0

C

_

_

The value of the load resistance RL is (A) R/4 (C) R (B) R/2 (D) 2R ECE - 2011 5. The circuit shown below is driven by a sinusoidal input . The steady state output is R

C

0.002

R

C

1mV +

(A) 0.5 A (B) 2.0A

(C) 1.0 A (D) 0.0A

ECE - 2007 3. Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2. The Value of C1

is

⁄ ⁄ ⁄ ⁄

(A) (B) (C) (D)

⁄ ⁄ ⁄ ⁄

ECE/EE/IN - 2013 6.

The transfer function

of the circuit

shown below is

L1

00μF

+ Vi

+ R

Vo

10k V1(s)

Filter 1

V2(s) 00μF

+ Vi

+ R

Vo

Filter 2

(A) 4 (B) 1

(C) (D)

(A)

(C)

(B)

(D)

⁄ ⁄ th

th

th


GATE QUESTION BANK

Network Theory

EE – 2014 1. The driving point impedance Z(s) for the circuit shown below is

F

F

(A)

(C)

(B)

(D)


[Ans. B]

2.

[Ans. A]

5.

[Ans. A] Redrawing the circuit s – domain

I(S) +

~

( ‖ )

LS V(S)

(

)

Li (0 )

+

…………… ⁄ 0

3.

0

So here, Now,

[Ans. D] ⁄

4.

0

⁄

⁄

Put ,

[Ans. C] (

(

(

⁄

(

⁄

So,

)

⁄ (

⁄

)

) ⁄ (

⁄

)

)) ⁄ (

+ ………………………

I(s) =

⁄ (

⁄

*

⁄

Now ,

)

( ‖ )

)

Satisfies above equation th

th

th


GATE QUESTION BANK

Network Theory

In time domain,

( )

6.

[Ans. D] 100 μF 0 100 μF

⁄ ⁄ Substituting the values we get

EE 1.

Ans. A]

F

peda e f

F

du a e

peda e f apa

‘ ’d

a

f F

(

)

th

th

th


GATE QUESTION BANK

Network Theory

Two-Port Networks ECE - 2006 1. In the two port network shown in the figure below, Z12 and Z21are, respectively

r

(A) r n (B) n

(C) n r (D) r n

r

The z-parameter matrix for this network is

4.

r

r

3.

*

+[

(C) *

+

(B) *

+

(D) *

+

(A) *

+

(C) *

(B) *

+

(D) *

Z(s)

. The component values are

]

(A)

(C)

(B)

(D)

ECE - 2008 Statement for linked Answer Questions 3 and 4 A two port network shown below is excited by external dc sources. The voltages and currents are measured with voltmeters and ammeters (All assumed to be ideal) as indicated. Under following switch conditions, the readings obtained are: (i) –open , - closed –closed ,

+

The driving point impedance of the following network is given by

If port-2 is terminated by RL, then input impedance seen at port-1 is given by

(ii)

+

r

A two-port network is represented by ABCD parameters given by [ ]

+

The h-parameter matrix for this network is

5. 2.

(A) *

s

(A) (B) (C) (D)

R

L

L H R Ω L H R Ω L = 0.1H, R = 2Ω L H R 2Ω

F F 0.1F F

ECE - 2010 6. For the two-port network shown below, the short circuit admittance parameter matrix is Ω

1

2 Ω

Ω

Ω 2

(A) *

– open

(B) *

2 2

+S

(C) * +S

(D) *

+S 2 2

+S

2 wo port n twork

ECE - 2011 7. In the circuit shown below, the network N is described by the following Y matrix:

2

* th

th

+. th


GATE QUESTION BANK

The voltage gain

s

is

Network Theory

s s

s s s

2

s

s s

s

s s

s

s s

s s s

N

s s s s s s

(A) 1/90 (B) 1/90

(C) (D)

1/99 1/11

11.

ECE/EE/IN - 2012 Common Data for questions 8 and 9 With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed: (i) 1 connected at port B draws a current of 3 A (ii) 2.5 connected at port B draws a current of 2 A

2

2 2

12.

8.

9.

In the h-parameter model of the 2-port network given in the figure shown, the value of h (in S) is ______ .

Consider the building block ‘N twork N’ shown in th figur L t F n R k N twork N

With 10 V dc connected at port A, the current drawn by 7Ω connected at port B is (A) 3/7 A (C) 1 A (B) 5/7 A (D) 9/7 A

called

R

Two such blocks are connected in cascade, as shown in the figure.

For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is (A) 6 V (C) 8 V (B) 7 V (D) 9 V

s

N twork N

N twork N

h tr nsf r fun tion ECE - 2014 10. A two-port

2

2

network

has scattering s s parameters given by [ ] *s s + If the port-2 of the two- port is short circuited, the s parameter for the resultant one-port network is th

of the cascaded

network is (A)

(C) (

(B)

(D)

th

th

s

)


GATE QUESTION BANK

13.

For the two-port network shown in the figure, the impedance (Z) matrix (in ) is

I1

Network Theory k

k

k

2

49i1

Input loop F F

2

(A) * (B) *

2 2

(C) *

+

2

(D) *

+

2 2

(A) 2 F (B) F

+ +

EE - 2006 1. The parameters of the circuit shown in the figure are R MΩ R = 10 Ω, A = V/V.If = 1 V, then output voltage, input impedance and output impedance respectively are R

R +

(C) 2 F (D) Fs

EE - 2010 4. The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d), respectively. It has an impedance matrix Z with parameters denoted by . A 1Ω resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified twoport network (shown as a dashed box) is

+ _

e

a

c

_

(A) 1V, , 10 Ω (B) 1 V,0, 10Ω 2.

P

(C) 1 V, 0, (D) 10 V, ,10Ω

f

The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are

d

b

(A) [

]

(B) [

]

(C) [

]

(D) [

]

IN - 2007 (A) z parameters,*

+

(B) h parameters,*

+

(C) h parameters,*

+

1.

The DC voltage gain circuit is given by. R

(D) z parameters,*

in the following R

R

+

EE - 2009 3. The equivalent capacitance of the input loop of the circuit shown is

th

(A) AV

(C) AV

(B) AV

(D) AV

th

th

+R


GATE QUESTION BANK

IN - 2008 2. For the circuit shown below the input resistance R11 =

|

3I2

2V3

Ω

+

I2 V3

2Ω

V1

+

2

2

1

2Ω

+I 1 V1

IN - 2013 3. Considering the transformer to be ideal, th tr nsmission p r m t r ‘ ’ of th 2 port network shown in the figure below is

is +

Network Theory

1:2

I1

5

5

V2

(C) (D)

I2

V2 2

(A) 1.3 (B) 1.4 (A) Ω (B) 2 Ω

2

(C) 0.5 (D) 2.0

Ω Ω


3.

[Ans. C] = =

[Ans. B] |

current source will be

s (1)

open)

|

=0

|

| |

2.

=0

=

r

s (2) |

[Ans .D] The ABCD parameter equations are given by,

=* 4.

wh n th n twork is t rmin l R R R

R

= 1.5 +

[Ans. A] =h h =h h From giv n p r m t rs 2

fig

R R

= 1.5 H=*

RL

= 3 +

Fig. 1

th

th

th


GATE QUESTION BANK

5.

Network Theory

[Ans. D] R

L R L ⁄

(R L

Note:

R ⁄ R L⁄

R L

is Independent of

L⁄ ) 8.

R

[Ans. C] As per the given conditions, we can draw the following two figures.

L⁄

RL R L

L

R

(

⁄R ⁄L ) 2 s 2 Comparing (1) and (2) 2

⁄ ⁄R ⁄L

A

B

± 10 V

2

F R

2

2

L

2A

H

B A

6.

3A

N

⁄

±

N

10V

2.5

[Ans. A] | | | [ ]

7.

*

Let Vth and Rth be the Thevenin voltage & resistance as seen from part B.

2 2 2

+

R

[Ans. D] iv n

*

+ 2

2 From the circuit shown in Fig. 1

R 2

2

N

Vth = 3Rth + 3 (1) & Vth = 2Rth + 5 (2) Solving (1) & (2) Rth 2 So, Vth = 3 x 2 + 3 = 9V Now,

Fig

From eqn (2)

th

th

th


GATE QUESTION BANK

Network Theory

i

N

2

N

i= 9.

The y parameter of the parallel network is equal to the sum of the individual network y parameter For network A

[Ans. B] R

2

So, Vth = 7/3 x 2 + =

= 7V.

The open circuit voltage at port B is 7V. 10.

2

[Ans. B]

p r m t rs

ort

ort 2

Lo

2 For network B

s s s Port 2 is short circuited h n ws

2 2

2

From s

s [

s ] s

s s

2

From s

s [

s s 11.

s

2

s ] s s [ s s s s s s

s s

[Ans. *] Range 1.24 to 1.26 In the figure two port networks in parallel

th

th

th


GATE QUESTION BANK

EE 1.

h

Network Theory

[Ans. A] = 1V =

2

2

= 12.

s

2.

s

[Ans. C] h h

s

V1

Apply mesh analysis to determine the current s )

s

(2R

s

s )

R s

s

Since port – 1 is open – circuit , I1=0 Port – 2 is short – circuit, V2 =0

s

)

s

[from

)

R]

s

)

R +

s

]

n

) (2R s R ) (R *(2R

s

R

s R s s

(2R

R

s

s

) (R

)

R s Rs

2Rs

k F s s

h

|

h

|

h

|

h

|

s

R s

o h – p r m t rs

R 3.

}R

h h

h ] h

*

+

k

k

i

s s

[

[Ans. A] Assume a 1A current source at input terminals, = 1A

R s sr s R R

V2

s

(2R

[(R

s

s

R

om ining

s

I2

s s

(R

h h

I1

R

R s

13.

R = 10

[Ans. B]

i

k

s

~

F F

[Ans. C]

i

[

]

*

Applying KVL i i [2

+ *

2

+

i

j ]

j

Input impedance

2

j

As imaginary part is negative, input impedance has equivalent capacitive reactance . th

th

th


GATE QUESTION BANK

Network Theory

2 h r

2

2

2

put it in qu tion

2 2 F 4.

o[

[Ans. C] The impedance matrix of the modified network is calculated from fig. given below:

3.

]

[Ans. A] 2

1 I1

[ ][ ]

[ ]

|

V1

2

5

2

5

2

V2

2

2 [

[ ]

n

2

|

][ ]

[

[ ]

]

*

t

+

2 2 From qu tion

1

2

2

From qu tion

Fig.

2 IN 1.

From qu tion

2 n

[Ans. A] V= =

2.

[Ans. D] 2 2V3

Ω + I1

+

f

2Ω

d

+ +

b

V1

3I2 I2

c 2

+

V3

2Ω

e

Apply KVL on abcdef

th

th

th


GATE QUESTION BANK

Network Theory

Network Topology ECE - 2008 1. In the following graph, the number of trees (P) and the number of cut-sets (Q) are

Which of the following statements is true? (A) The equations v1 v2+v3 = 0, v3 +v4 –v5 =0 are KVL equations for the network for some loops (B) The equations v1-v3-v6 = 0, v4 +v5 –v6 =0 are KVL equations for the network for some loops (C) E = AV (D) AV = 0 are KVL equations for the network

(1)

(2)

(3) (4)

(A) P = 2, Q = 2 (B) P = 2, Q = 6

(C) P = 4, Q = 6 (D) P = 4, Q = 10

EE - 2007 1. The matrix A given below is the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes. Let V = [v1 v2… v6]T denote the vector of branch voltages. While I = [i1 i2… i6]T that of branch currents. The vector E=[e1 e2 e3 e4]T denotes the vector of node voltages relative to a common ground. [

EE - 2008 2. The number of chords in the graph of the given circuit will be

±

(A) 3 (B) 4

(C) 5 (D) 6

]

th

th

th


GATE QUESTION BANK

Network Theory


The graph of the network is shown in fig. Where From the graph it can be observed that (i) are not KVL equations as set of branches (1, 2, 3) and (3, 4, 5) do not form closed paths. (ii) and are KVL equations for the loops (1, 3, 6) and (4,5, 6) From the matrix, A it can be concluded that (i) E ≠ (ii) AV = 0 are not KVL equations Statement in option (b) is true.

[Ans. C] Different trees (P) are shown below.

Different cut sets (Q are shown below

(1)

(2)

(3)

(5)

(6) (4)

So P = 4, Q = 6 2. EE 1.

[Ans. A] The graph of the given circuit is shown in Fig. Number of nodes = N = 4 Number of branches = B = 6 Number of tree branches = (N – 1) = 3 Number of links = L = B – (N – 1) = 3

[Ans. B] For the given node – to – branch incidence matrix [

] 1 5

2

I

III

II

Fig. 4

3

IV

6

Fig.

th

th

th


GATE QUESTION BANK

Signals and Systems

Introduction to Signals & Systems ECE - 2006 1. The Dirac delta function t is defined as t (A) t , t rw s t (B) t , t rw s t (C) t , t rw s ∫ (D)

t

, ∫

t t t t

rw s

t t

ECE - 2007 2. The 3-dB bandwidth of the low-pass signal u t , where u(t) is the unit step function, is given by (C) (A) z (D) z √√ (B) z ECE - 2009 3. A function is given by f(t) = sin2t +cos2t . Which of the following is true? (A) f has frequency components at 0 and / π z (B) f has frequency components at 0 and /π z (C) f s fr qu cy c mp ts t / π /π z (D) f has frequency components at 0, / π /π z ECE - 2011 4. If the unit step response of a network is , then its unit impulse response is (A) (C) (B) (D)

ECE /EE/IN- 2013 5. Two systems with impulse responses (t) and (t) are connected in cascade. Then the overall impulse response of the cascaded system is given by (A) Product of h1(t) and h2(t) (B) Sum of h1(t) and h2(t) (C) Convolution of h1(t) and h2(t) (D) Subtraction of h2(t) from h1(t) ECE - 2014 6. A discrete-time signal x[n] s π t (A) Periodic with period π (B) Periodic with period π (C) Periodic with period π/ (D) t p r c

r s

7.

A system is described by the following differential equation, where u(t) is the input to the system and y(t) is the output of the system. ẏ t y t u t When y (0) = 1 and u(t) is a unit step function, y(t) is (A) (C) (B) (D)

8.

The sequence x[n] = u[n] , where u[n] is the unit step sequence, is convolved with itself to obtain y[n]. Then ∑ y is___________

EE - 2006 1. Which of the following is true: (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [ t , t ] for some t (D) A bounded signal is always finite

th

th

th


GATE QUESTION BANK

EE - 2007 2. If u(t), r(t) denote unit the unit step and unit ramp functions respectively and u(t) * r(t) their convolution, then the function u (t+1) * r(t 2) is given by (A) (1/2) (t 1) (t+2) (B) (1/2) (t 1) (t 2) (C) (1/2) (t 1)2 u(t 1) (D) none of the above

Signals and Systems

EE - 2014 5. The function shown in the figure can be represented as

t

EE - 2008 3. Given a sequence x[n], to generate the sequence y[n] = x[3 4n], which one of the following procedures would be correct?. (A) First delay x[n] by 3 samples to generate z [n], then pick every 4th sample of z1[n] to generate z2[n], and then finally time reverse z2[n] to obtain y[n] (B) First advance x[n] by 3 samples to generate Z1[n], and then pick every 4th samples of z1[n] to generate z2 [n] and then finally time reverse Z2[n] to obtain y[n] (C) First pick every fourth sample of x[n] to generate v time-reverse v to obtain v & finally advance v by 3 sample to obtain y[n] (D) First pick every fourth sample of x[n] to generate V1[n], time-reverse V1[n] to obtain V2[n], and finally delay V2[n] by 3 samples to obtain Y[n] EE - 2011 4. A zero mean random signal is uniformly distributed between limits and its mean square value is equal to its variance. Then the r. m. s. value of the signal is (A) (C) √ √ (D) √ (B) √

(A) u t

u t

u t

u t (B) u t (C) u t

u t

u t

u t

u t

u t (D) u t

u t

u t

6.

An input signal x t s πt is sampled with a sampling frequency of 400 Hz and applied to the system whose transfer function is represented by z z ( ) z Where, N represents the number of samples per cycle. The output y(n) of the system under steady state is (A) 0 (C) 2 (B) 1 (D) 5

7.

For the signal f t s πt s πt s πt the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is ________

IN - 2007 1. Consider the periodic signal x t c s πt c s πt where t is in seconds. Its fundamental frequency, in Hz, is (A) 20 (C) 100 (B) 40 (D) 200 th

th

th


GATE QUESTION BANK

IN - 2008 2. The fundamental period of the discretetime signal x[n] = (A) ⁄ π ⁄ (B) 3.

(

The integral

is (C) 6 (D) 12

The step response of a linear time invariant system is y t u t , where u(t) is the unit step function. If the output of system corresponding to an impulse input t is h(t), then h(t) is (A) u t (B) t (C) u t u t (D) t u t

π ) s

∫ (t

)

IN - 2009 4. The fundamental period of x(t) = 2sin2πt +3sin3πt , with t expressed in seconds, is (A) 1s (C) 2s (B) 0.67s (D) 3s 5.

7.

Signals and Systems

t

t v u t t

(A) 6 (B) 3

(C) 1.5 (D) 0

IN - 2011 8. Consider a system with input x(t) and output y(t) related as follows y(t) =

{

x t }

Which one of the following statements is TRUE? (A) The system is nonlinear (B) The system is time-invariant (C) The system is stable (D) The system has memory 9.

The continuous-time signal x(t) = s t is a periodic signal. However, for its discrete-time counterpart x[n] = s to be periodic, the necessary condition is (A) ≤ π

For input x(t), an ideal impulse sampling system produces the output

(B)

to be an integer

(C)

to be a ratio of integers

y t

(D) none

∑ x

t

Where t is the dirac delta function. The system is (A) Non-linear and time invariant (B) Non-linear and time varying (C) Linear and time invariant (D) Linear and time varying IN - 2010 6. The input x(t) and the corresponding output y(t) of a system are related by . y(t) = ∫ (A) (B) (C) (D)

x

. The system is

EC/IN - 2013 10. If the A-matrix of the state space model of a SISO linear time invariant system is rank deficient, the transfer function of the system must have (A) A pole with a positive real part (B) A pole with a negative real part (C) A pole with a positive imaginary part (D) A pole at the origin 11.

time invariant and causal time invariant and non-causal time variant and non-causal time variant and causal

th

The impulse response of a continuous time system is given by h(t) = t t . The value of the step response at t = 2 is (A) 0 (C) 2 (B) 1 (D) 3

th

th


GATE QUESTION BANK

IN - 2014 12. Time domain expressions for the voltage t t r v s t s t t = c s t Which one of the following statement is TRUE?

(A) (B) (C) (D)

Signals and Systems

t t t t

s

t y t y t y t y

s s s


s Every after s

[Ans. D]

t

π s π trigonometric function repeats π t rv π π s π π

π

) π S c ‘ ’ s y teger, there is no p ss v u f‘ ’f rw c ‘ ’c an integer, thus non-periodic

t 2.

[Ans. A] f m

tu

√

s

w

7.

s t

fr qu

cy

[Ans. A] ẏ t y t I.F. =

√

y t

w

√

r f

3.

u t

8. c s t

π

c s t

π

π

∑ y

∑ x

Here x

z

∑ x

t ∑ u

∑

[Ans. A] t

6.

∫

[Ans. *] Range 3.9 to 4.1 We know, if y x

Frequency components are f

5.

u t , P = 5, Q = u t ,

y t

π

f t

4.

(

y

[Ans. B]

f

π

t

s

y t

[Ans. C ] Cascade means convolution [Ans. D] Assume x N) x x

EE 1.

∑ y

[Ans. D] If the amplitude of a signal have some finite boundaries for all values of time then it is called as bounded signal.

to be periodic, (with period

th

th

th


GATE QUESTION BANK

|f t | }f r t r |f t | f tt v v u So a bounded signal may possess finite energy or infinite energy. For example u (t) is bounded signal but it possess infinite energy because it is a power signal. It can be zero or nonzero outside a finite interval ( t t But it is always true that it will be always finite for any value of time t.

Signals and Systems

i.e, z z x Now reverse (time reverse) z give y[n] = z x 4.

will

[Ans. A] Variance

Mean square value 2.

[Ans. C] u t

RMS value

r t ∫r

z t

5.

u t

√

[Ans. A] Result graph

r p r p

r p

∫ r

r p u

z t

u t

∫

For (t

r p

≥ t

z t

(

u t

)

r p

f r t≥ t z t

t

For

t

t

t

t≥

z t

u t

t

For

t≥

z t 3.

t

t

u t

6.

[Ans. C] x t t πft

[Ans. B] Y[n] = x[3 – 4n] = x[ 4n+ 3] So to obtain y[n] we first advance x[n] by 3 unit. i.e, z x Now we will take every fourth sample of z

m p r

s πt

πt f

fx t ms

s mp um th

th

p r

r f s mp s th

ms c cyc


GATE QUESTION BANK

Signals and Systems

Now, fundamental period of x(t) is the LCM of T1, T2 and T3 , which is x z

So, fundamental frequency = 20 Hz

s mp p r r st y st t my m m [

z

z

2. z

y z

IN 1.

[Ans. D]

z x z

x Fundamental period

z πt πt

3.

]

[Ans. D] Step response s(t)

u t

Impulse response h(t)

z

zs zc s

z + z (Using L Hospital rule) m*

m

z z zs zc s

z z z

m*

7.

s

πt πt z m z

{

t +

[Ans. 14] r qu cy f s πt z r qu cy f s πt z r qu cy f s πt z By Nyquist theorem, sampling frequency f z x mum fr qu cy f z

u t u t 4.

u t / t t t

[Ans. C] x t S πt S P r f S πt s P r

f

S

πt

πt s ⁄

P r of final original is LCM of = LMC / = 2 sec. 5.

[Ans. D] Given signal is y t

[Ans. A] X(t) = (1+0.5 cos πt c s πt c s πt c s πt c s πt c s πt c s πt c s πt x t x t x t w r x c s πt x c s πt x c s πt Now, tm p r fx t π s π

}

∑ x

t

This is the representation of a signal x(t) in weighted and sum form and it obeys the principle of superposition and homogeneity. So, this is linear but it is time varying as for x(t t ), y t ∑x t t And y t t ∑x t t Therefore y t t y t s t m v ry 6.

[Ans. C] x t

y t

∫ x

s Value of y t ‘t’ depends on values of x for times from t t As output is depending on future values of input, the system is Non casual

s

th

th

th


GATE QUESTION BANK

For the input x t y t

∫ x

y t

∫ x

x t

t

As y(t) depends only on x(t), the system has no memory. 9.

t

[Ans. C] If x(n) is periodic with period N, then the condition to be satisfied is x(n) = x(n+N). If x(n) = s then the necessary condition is s s Or N= m π m 1, 2,

t

From (1), y(t-t

x

∫

Let From (3),

t

Or

y t

x ∫ As (4) & (5) are not same, the system is Time variant. syst m s m v r t causal. 7.

∫

r t

Or

[Ans. D]

11.

[Ans. B] Y(s)

π

(t

t

t

rs

(step response)

u t ) s

f

should be expressible as π ( )

10.

[Ans. B] Given signal is x t

Signals and Systems

t

u t

u t

u t

By shifting property of unit impulse function ∫x t ∫

t (t

t π

x t {

t

) s

t

t

t

t t s w r s

u t

u t

π At t = 2 value is 1

Hence, correct option is (B) 8.

12.

[Ans. C]

[Ans. A] t

Given the system, y(t) = y t

t

x t

{

t

x t }

x t

[

x t x t ] t As y(t) is obtained from linear operation on x(t), the system is linear . As the input x(t) is multiplied by a time varying function , the system is Time – varying. For a bounded input x(t) , x(t) is bounded, y(t) is also bounded. Therefore the system is stable. th

v t t t

t

t

s

c s c s c s

th

s t t t

t

t y

th


GATE QUESTION BANK

Signals and Systems

Linear Time Invariant (LTI) systems ECE - 2006 1. A system with input x given

as

y

(s

5. and output y π )x

.

Let x(t) be the input and y(t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4. Properties Relations P1: Linear but NOT : y(t) = t x (t) time-invariant P2: Time-invariant : y(t) = t |x (t)| but NOT linear P3: Linear and : y(t) = |x (t)| time-invariant :y t x t 5) (A) (P1, R1), (P2, R3), (P3, R4) (B) (P1, R2), (P2, R3), (P3, R4) (C) (P1, R3), (P2, R1), (P3, R2) (D) (P1, R1), (P2, R2), (P3, R3)

is The

system is (A) linear, stable and invertible (B) non-linear, stable and non-invertible (C) linear, stable and non-invertible (D) linear, unstable and invertible ECE - 2008 2. A discrete time linear shift-invariant system has an impulse response h[n] with , and zero otherwise. The system is given an input sequence x[n] with x[0] = x[2] =1, and zero otherwise. The number of nonzero samples in the output sequence y[n] and the value of y[2] are, respectively (A) 5, 2 (C) 6, 1 (B) 6, 2 (D) 5, 3 3.

4.

ECE - 2009 6. Consider a system whose input x and output y are related by the equation :

The impulse response h(t) of a linear time-invariant continuous time system is described by h(t) = exp( t)u(t) + exp( t)u( t), where u(t) denotes the unit step function, and and are real constants. This system is stable if (A) is positive and is positive (B) is negative and is negative (C) is positive and is negative (D) is negative and is positive

y t

∫ x t H(t)

0

(t)

Where h(t) is shown in the graph. Which of the following four properties are possessed by the system? BIBO: Bounded input gives a bounded output Causal: The system is casual LP : The system is low pass LTI : The system is linear and timeinvariant (A) Casual ,LP (B) BIBO ,LTI (C) BIBO, Casual, LTI (D) LP , LTI

The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following description corresponds to a causal system? (A) y(t) = x(t 2) + x (t + 4) (B) y(t) = (t 4) x (t + 1) (C) y(t) = (t + 4) x (t 1) (D) y(t) = (t + 5) x (t + 5)

th

th

th


GATE QUESTION BANK

Signals and Systems

ECE - 2010 7. Two discrete time systems with impulse responses and are connected in cascade. The overall impulse response of the cascaded system is (A) (B) (C) (D)

ECE - 2013 12. The DFT of a vector c is the vector . Consider the product c c pqrs c [ ]. c c The DFT of the vector p q r s is a scaled version of (A)

ECE - 2011 8. A system is defined by its impulse response u . The system is (A) Stable and causal (B) Causal but not stable (C) Stable but not causal (D) Unstable and noncausal

ECE - 2014 13. The value of the ∫ s c t t is ________.

9.

An input is applied response (A) (B) (C) (D)

x t xp t u t t to an LTI system with impulse t u t The output is xp t u t u t xp t u t u t xp t u t u t xp t u t u t

(B) [√ √ √ √ ] (C) (D)

14.

A continuous, linear time-invariant filter has an impulse response h(t) described f r ≤t≤ by t , t rw s When a constant input of value 5 is applied to this filter, the steady state output is_____.

15.

Consider a discrete-time signal f r ≤ ≤ x , t rw s If y[n] is the convolution of x[n] with itself, the value of y[4] is _________.

16.

The input-output relationship of a causal stable LTI system is given as y y x If the impulse response of this system satisfies the condition ∑ , the relationship between and is (A) / (C) (B) / (D)

17.

Let h(t) denote the impulse response of a

ECE /EE/IN- 2012 10. The input x(t) and output y(t) of a system are related as y(t) = ∫ x c s The system is (A) time-invariant and stable (B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable 11.

Let y[n] denote the convolution of h[n] and g[n], where h[n]= (1/2)n u[n] and g[n] is causal sequence. If y[0]=1 and y[1]=1/2, then g[1] equals (A) 0 (C) 1 (B) 1/2 (D) 3/2

integral

causal system with transfer function

.

Consider the following three statements. S1: The system is stable. th

th

th


GATE QUESTION BANK

S2:

y [ 0 ] = 3, y[ 2 ] = 2 is the output of a discrete – time LTI system. The system impulse response h [ n ] will be (A) h [ n ] = 0; n < 0, n > 2, h[ 0 ] = 1, h[1]=h[2]= 1 (B) h [ n ] = 0; n < 1, n > 1, h[ 1 ] = 1, h[0]=h[1]=2 (C) h [ n ] = 0; n < 0, n > 3, h[ 0 ] = 1, h [ 1 ] =2, h [ 2 ] = 1 (D) h [ n ] = 0; n < 2, n > 1, h[ 2 ] = h [ 1 ] =h [ 1] = h[0]=3

is independent of t for t > 0.

S3: A non-causal system with the same transfer function is stable. For the above system, (A) only S1 and S2 are true (B) only S2 and S3 are true (C) only S1 and S3 are true (D) S1, S2 and S3 are true 18.

A real-valued signal x(t) limited to the frequency band |f| ≤ is passed through a linear time invariant system whose frequency response is |f| ≤ f

{ |f|

The output of the system is (A) x t (C) x t (B) x t (D) x t 19.

A stable linear time invariant (LTI) system has a transfer function H(s)

. To make this system causal

it needs to be cascaded with another LTI system having a transfer function s . A correct choice for s among the following options is (A) s (C) s (B) s (D) s EE - 2006 1. A continuous – time system is described by y(t) = | | , where y(t) is the output and x(t) is the input. y(t) is bounded (A) only when x(t) is bounded (B) only when x(t) is non – negative (C) only for t ≤ 0 if x(t) is bounded for t≥0 (D) even when x(t) is not bounded 2.

x[n] = 0; n < 1, n > 0, x [ 1] = 1, x[ 0 ] = 2 is the input and y[ n ] = 0; n < 1, n > 2, y [ 1 ] = 1 = y [1],

Signals and Systems

EE - 2007 3. X(z) = 1 z , Y(z)= 1+ z are Z – transforms of two signals x[n], y[n] respectively. A linear time invariant system has the impulse response h[n] defined by these two signals as h[n] = x[n 1] y[n] where denotes discrete time convolution. Then the output of the syst m f r t put -1] (A) Has Z-transform X(z) Y(z) (B) qu s 2] 4] 5] (C) s tr sf rm z z z (D) Does not satisfy any of the above three. EE - 2008 4. A signal sin ( t) is the input to a Linear Time Invariant system. Given K and are constants, the output of the system will be of the form sin (vt ) where (A) t qu t ; but equal to (B) v need not be equal to ut qu to (C) qu t and equal to (D) t qual to and need not be equal to

th

th

th


GATE QUESTION BANK

5.

6.

The impulse response of a causal linear time-invariant system is given as h(t). Now consider the following two statements Statement (i): Principle of superposition holds Statement (ii) : h(t) = 0 for t < 0 Which one of the following statements is correct? (A) Statement (i) is correct and Statement (ii) is wrong (B) Statement (ii) is correct and Statement (i) is wrong (C) Both Statement (i) and Statement (ii) are wrong (D) Both Statement (i) and Statement (ii) are correct A system with input x(t) and output y(t) is defined by the input-output relation, y (t) = ∫

The system will be

(A) causal, time-invariant and unstable (B) causal, time-invariant and stable (C) non – causal, time-invariant and unstable (D) non - causal, time-variant and unstable EE - 2009 7. A Linear Time Invariant system with an impulse response h(t) produces output y(t) when input x(t) is applied. When the input x (t s pp t system with impulse response h( ), the output will be (A) y(t) (C) y(t (B) y(2(t (D) y(t 8.

The z transform of a signal x[n] is given by z + z + 2 – 6z2 + 2z3. It is applied to a system, with a transfer function H(z) = z 2. Let the output be y[n]. Which of the following is true? (A) y[n] is non causal with finite support (B) y[n] is causal with infinite support

Signals and Systems

(C) y[n] = 0;|n|>3 (D) Re[Y[z] = Re[Y[z] Im[Y[z] = Im[Y[z] π≤ θ π 9.

;

A cascade of 3 Linear Time Invariant systems is causal and unstable. From this, we conclude that (A) Each system in the cascade is individually causal and unstable (B) At least one system is unstable and at least one system is causal (C) At least one system is causal and all systems are unstable (D) The majority are unstable and the majority are causal

EE - 2010 10. Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is h[n] y[n] = {1, 0, 0, 0, 1}

x[n] = {1, 1}

{

(A)

{

(B)

}

{

(C)

}

{

(D) 11.

}

}

The system represented by the inputoutput relationship y t (A) (B) (C) (D)

t is ∫ x Linear and causal Linear but not causal Causal but not linear Neither linear nor causal

EE - 2011 12. The response h(t) of a linear time v r t syst m t mpu s t under initially relaxed condition is th

th

th


GATE QUESTION BANK

13.

h(t) = . The response of this system for a unit step input u(t) is (A) u(t) + (B) u t (C) u t (D) t u t Given two continuous time signals x(t) = and y(t) = which exist for t > 0 the convolution z(t) = x(t) * y(t) is (A) u t (B) u t (C) (D)

EE - 2013 14. Which one of the following statement is NOT TRUE for a continuous time causal and stable LTI system? (A) All the poles of the system must lie on the left side of the j axis. (B) Zeros of the system can lie anywhere in the s-plane (C) All the poles must lie within |s| (D) All the root of the characteristic equation must be located on the left side of the j axis.

If the input to the system is cos(3t) and the steady state output is A sin(3t + ), then the value of A is (A) 1/30 (C) 3/4 (B) 1/15 (D) 4/3 17.

Consider an LTI system with impulse response h(t) = u t . If the output of the system is y t u t u t then the input, x(t), is given by (A) u t (B) u t (C) u t (D) u t

18.

A 10 kHz even-symmetric square wave is passed through a bandpass filter with centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is (A) a highly attenuated square wave at 10 kHz. (B) nearly zero. (C) a nearly perfect cosine wave at 30 kHz. (D) a nearly perfect sine wave at 30 kHz.

19.

A continuous-time LTI system with system function H( ) has the following pole-zero plot. For this system, which of the alternatives is TRUE?

EE - 2014 15. x(t) is nonzero only for t , and similarly, y(t) is nonzero only for t . Let z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE? (A) z(t) can be nonzero over an unbounded interval. (B) z(t) is nonzero for t (C) z(t) is zero outside of t (D) z(t) is nonzero for t . 16.

(A) | (B) | (C) | (D) |

Consider an LTI system with transfer function s

Signals and Systems

| | | |

| | | | s mu t p m x m | | | | c st t

t

s s

th

th

th


GATE QUESTION BANK

IN - 2007 1. The signals x(t) and h(t) shown in the figures are convolved to yield y(t) .

3.

Consider a discrete-time LTI system with input x [n] = [n] + [n 1] and impulse response h[n] = [n] – [n 1]. The output of the system will be given by (A) [n] – [n 2] (B) [n] – [n 1] (C) [n 1]+ [n 2] (D) [n]+ [n 1]+ [n 2]

x(t) 1

1 1

t

1 h(t)

1 0

4

2

t

Which one of the following figures presents the output y(t)? (A) y(t)

IN - 2009 4. A linear time-invariant casual system has frequency response given in polar form as:

1

t

√

1

3

1

1

For input x(t) = sint, the output is

t

7

5

(A) (B)

(B) y(t)

(C)

1 4

(C)

6

2

1

(D)

t

y(t)

1 1

Signals and Systems

3

1

t

5

√ √ √ √

c st c s (t s t s

(t

)

IN - 2011 5. Consider the signal t≥ x(t) = { t Let X( ) denote the Fourier transform of this signal. The integral is

∫ (D)

)

(A) 0 (B) ⁄

y t) 1 3 1

2

5

t

(C) 1 (D)

IN - 2013 6. The impulse response of a system is h(t) = t u(t). for an input u(t 1), the output is (A)

IN - 2008 2. Which one of the following discrete-time systems is time invariant? (A) y [n] = nx [n] (C) y[n] = x[ n] (B) y [n] = x [3n] (D) y [n] = x[n 3]

u t

(B)

u t

(C)

u t

(D)

th

u t

th

th


GATE QUESTION BANK

Signals and Systems

IN - 2014 7. The impulse response of an LTI system is given as : {s

π

π It represents an ideal (A) non-causal, low-pass filter (B) causal, low-pass filter (C) non-causal, high-pass filter (D) causal, high-pass filter


[Ans. C] x

→

x[k]

y

[s

( π )] x

The system is linear as y(n) is proportional to the input x(n) Times varying as x(n) is multiplied not by a constant but by a function of n. Stable as bounded input, x(n) gives bounded output , y(n) : |s ( π )| ≤

and |y

| ≤ |x

2 1

0 1

h[k]

|

+1 2

0

Not invertible as several inputs: k (n) , where k is any arbitrary integer, give the same output equal to zero. The system is Linear, stable and non – invertible (It is also Time – varying). 2.

1

1

1

[Ans. D] y

∑ x y y y y y y y

3.

[Ans. D] h(t) = u t u t For the system to be stable , ∫

t t

For the above condition, h(t) should be as shown below.

th

th

th


GATE QUESTION BANK

Signals and Systems

t

∑ u t

u t

9.

s

v

syst m s

[Ans. D]

t

s

s

s y(s) = X(s).H(s)

Therefore Graph for

s

s s

y t [Ans. C] A system is casual if the output at any time depends only on values of the input at the present time and in the past.

5.

[Ans. A]

6.

[Ans. B] y t

10.

x t

y t

∫ x

c s

check for time invariant ∫x

y t f r

∫x t

c s

t y

utput

∫ x

c s

For delayed input

t

y t

It is not low pass filter. But the system is LTI and BIBO.

8.

u t

[Ans. D]

y t

∫x t

t

c s

P p

[Ans. C]  H(z) = 

s

u t

h(t)

7.

s

s

s s taking inverse w t

4.

t st

z

where t P t t When P Also P t

z =

[Ans. B] u For causal system h(n) = 0 for n < 0. Hence given system is casual. For stability:

y t

∫ x P c s

P

t

p

s c y t y t t It is not time invariant for a bounded input, x (t) = cos (3t) u(t) th

th

th


GATE QUESTION BANK

Signals and Systems

x y t

∫c s

{x

x

} { c um v ct r }

x

x ∫ su

11.

{x x Designing the sequence x x x x

∫ c s u syst m s

st t stable.

[Ans. A]

g(1) g(0)

∑ 1[- k] 1/2

1

1/4 k

0 1 0 1 c c y

x

x x x x x [ x ] Now circular convolutions given by x x x c c [ ] c c Now circular convolution of c with it self c c c c [ ]0 1 c c c Taking transpose both side c c 0 1 0 1 c [ ] c c c c Let

( ) u

y

} { c um v ct r } x matrix using same

∑

P

S

Take DFT both side

y P

y

0 1 0 1 c c

c

c

∑

y h[1 – k] will be zero for k > 1 and g[k] will be zero for k<0 as it is causal sequence.

P 13.

g [1] = 0 12.

S

[Ans. *] Range 0.19 to 0.21 f s c t r ct ( ) | | ∫ s c

[Ans. A] Matrix method

| | th

S

th

t

t

f ∫ |r ct ( )|

th

f

P rs v

st


r m

GATE QUESTION BANK f tr ( )

Signals and Systems

z

r

z

f

r

(∫ tr ( ) f)

∑

∑s

f ∫ tr ( ) f

| |

17.

[Ans. A] t

14.

[Ans. *] Range 44 to 46 f r ≤t≤ t , t rw s t u t u t s

*

s

y s

s

r st

s y st t

s

+

s S : Syst m s st c us t s ut y t S : s p t

t s

s

p

s

utput

u

m sy s

s c s t f rm f % s Hospital Rule?

pp y

‘ ’ 18.

t ftm t ftm

x t

[Ans. *] Range 9.9 to 10.1 f r ≤ ≤ x , t rw s x z z z z z Convolution in time domain is multiplication z domain z x z x z f rf y ,we have to find coefficient of z fy z S t s

t (a non-causal system)

[Ans. D] t

y t

[Ans. A] y y y z y z z y z x z z

t r

but this is not absolutely integrable thus unstable Only S and S are TRUE

m

16.

u t

S : A non-causal system with same transfer function is stable

s

m

15.

t

x t f x f f y t x t

19.

y t

t c v ut f f

[Ans. B] s

s s s s For a stable causal LTI systems, all system poles should lie on LHP of s-plane. So to make H(s) causal, we need to cancel out the pole at s = 2 So, s s

x x z

th

th

th


GATE QUESTION BANK

EE 1.

2.

= output of the LTI system due to the input = output due to the input

[Ans. D] y(t) = | | , |x t | is always positive for positive as well as negative values of x(t). When x(t) is bounded, y (t) is bounded. Even when x(t) is not bounded i.e, x(t) = ± f r y t Y(t) = y t s u v w x t s t bounded. [Ans. A] For finite duration convolution x[n] have M terms h[n] have N terms then, y[n] should have terms (M+N 1) Here, x[n] = { 1, 2}

Note that the following definitions and properties are used: z

y[n] = { 1, 3, 1, 2} so option (A) is correct . 3.

∑

x

z

If the input to an LTI system is shifted by units, the output is also shifted by units. 4.

[Ans. A] If the input x(t) = s t to a real Linear Time Invariant system with frequency response , then the output y(t) = s t If the input is a damped sinusoid need not equal to alpha but frequency v

5.

[Ans. D] An LTI system is causal if the impulse response , h(t) =0 for t< 0 The principle of superposition holds for a linear system. t st t m ts r c rr ct

6.

[Ans. D]

y[n] = { 1, 3, 1, 2} So, h[n] should have only 3 terms and h[n] have value starting from origin of [n=0] only because y[n] have start from [n = 1] s r pt 1 -1 1 1 1 1 1

Signals and Systems

Given y(t) = ∫

x

y at time t is depending on values of x(t) in the range of t = 0 to ( 2t). i.e., y for negative values of time is depending on x at positive values of time. Hence System is non – causal For a bounded input like a step function the output is not bounded , as the input is integrated.

[Ans. B] For (x) = 1 – 3z x(n) delaying by 1 x For Y(z) = z y x y

r

put x t

For x t th

th

utput y t

∫x t

t

t th


GATE QUESTION BANK

utput

t

∫ x t

∫ x t

greater than zero even so atleast one of them must be causal. Similarly even if one (alteast) of the system become unstable then overall system will become unstable.

t

t 10.

Where t But y t

t

t

x t t ∫ As equations (1) and (2) are not equal , System is Time – variant Syst m s – causal , Time – variant , and unstable. It may be noted that the system is Linear. 7.

8.

t

[Ans. D] Case 1: Y(s) = H(s). X(s) Case 2: input = x(t – s Impulse response = h(t – Output: Y(s) = X(s). . H(s) = X(s). H(s). y(t 2 [Ans. A] y[n] = x[n] *H[n] y[z]=x[z] H[z] y[z] = z z

Since output has

z

z

z

z z

z

z

z z

comparing with y[n] = [1, 0, 0, 0, - 1] z

z z

z

y f r Therefore it is not causal with finite support 9.

no. of elements

Where N is number of elements in impulse response h[n] –1 N=4 Let it be Then X(z)= z z z z y z x z z y z z z z z So, y

s

z

[Ans. C] x[n] = {1,-1} , M=2 y[N] = { 1,0,0,0, - 1} . N1 =5

z z

Signals and Systems

[Ans. B] Since in cascade overall impulse response h(t) = h1(t)* h2(t)* h3(t) h1(t), h2(t), h3(t) are impulse responses of individual systems. Since initial point where h(t) is nonzero is t ≥ 0 and since in convolution initial point t t t Where t t t are initial points of t t t respectively. So for it to be greater than zero atleast one of them t t t must be +ve

{

11.

}

[Ans. B] Integrator is always a linear support Since y(t) = ∫

x

For t = 1 y t

∫x

Here value at t = 1 depends on future v u s tt f put x t So it is a non-causal system.

th

th

th


GATE QUESTION BANK

12.

[Ans. C] For given LTI system with input response h(t) = t

15. t

Signals and Systems

[Ans. C]

, x t

Note that u(t) = ∫ t t. For an LTI system, if the input is integrated, the output is also integrated. r u t st p put u t t response, s(t) is given s t

∫

s t

t

y t

t t

∫

t

t

|

Thus z t is zero outside t by convolution

f r t s t

[Ans. B]

[Ans. A]

s

X(s) = s

s s

s s

|

s

|

√ |

t

|

√

Z(s) = taking partial fraction z t 14.

of

u t 16.

13.

property

u t

17.

[Ans. C] Since it is a stable system all poles should lie in the left half of s-plane |S| s ± This implies one pole lie in RHS

[Ans. B] t s

u t s

y s

s

y s s

s

x s

s s

s

s

t

th

s s

s

18.

s s

u t

[Ans. C] 10 kHz even symmetric square wave have frequency component present th

th


GATE QUESTION BANK

19.

Signals and Systems

10kHz, 30kHz, 50kHz, 70kHz [Only odd harmonics due to half wave symmetry] Since bandpass filter is centered at 30kHz, 30kHz component will pass through filter output is nearly perfect cosine wave at 30 kHz Cosine in due to reason that signal in even signal.

y(t)

[Ans. D]

-1

h(t+1)

1 1 1

t

5

3

21

h(t 1) Fig. 3 y(t)

1 2 2

1

-1

t

5

3

Fig. 4

x(t) = 1 t t y t t * t t * t using the properties of convolution: t * t t t * t t y t t t The resultant of the triangles in Fig. 3 is shown in Fig. 4

s (

)(

( ( (

su st tut w t| IN 1.

)(

)(

)

)(

)(

)(

)

)(

)

s |

c

st

t

)

2.

[Ans. D] y(n) =x(n 3) represents a time delay system with delay = 3 secs. It is a Time invariant system as can be confirmed from the following Test. Test for time – invariance: If y1(n) = y(n-n0), then the system is time invariant rt v syst m x y x -3)

[Ans. D] y(t) = x(t) * h(t) x(t) and h(t) are shown in Fig. 1 and Fig.2 x(t) 1 1 1 1

x x From (1) y

Fig. 1

y x x

h(t)

y

1 0

2

4

t

3.

[Ans. A] z z y

Fig. 2

th

th

z z

z z

z z

th


GATE QUESTION BANK

4.

Signals and Systems

[Ans. D]

T/F

x t

y t

Given, Transfer function

t

√

And input x(t) = sint t

√

y t 5.

s

√

√

t

[Ans. C] Given x (t) =

t≥ t

x t f r pr p rty: ∫

6.

f f

π

[Ans. C] t

tu t

⁄p

s y s y t 7.

x t |

∫

s

⁄s s *

u t s

s +

ut

[Ans. A] For a system to be non-causal its impulse response h(n) should depends upon future value of n and if h(n) is independent of the past value it never means that system is non-causal And it is the equation of LPF with cutoff frequency

th

th

th


GATE QUESTION BANK

Signals and Systems

Fourier Representation of Signals ECE - 2006 1. t x t ) be Fourier Transform pair. The Fourier Transform of the signals x(5t 3) in terms of X(j ) is given as

(A) (B)

X( ) X( )

(C)

X( )

(D)

X( )

ECE - 2012 6. The Fourier transform of a signal h(t) is H{j𝜔) = (2 cos𝜔) (sin 2 𝜔) /𝜔. The value of h(0) is (A) 1/4 (C) 1 (B) 1/2 (D) 2 ECE- 2013

ECE - 2007 2. A 5-point sequence x[n] is given as x[ 3]=1, x[ 2]=1, x[ 1]=0, x[0]=5, x[1]=1. Let X( denote the discretetime Fourier transform of x[n]. The value of ∫ ( ) s

(A) 5 (B) 10π

(C) 16π (D) 5+j10π

ECE - 2008 3. The signal x(t) is described by f r ≤t≤ x(t) =, . t rw s Two of the angular frequencies at which its Fourier transform becomes zero are (A) π π (C) 0, π (B) π π (D) π π ECE - 2009 4. The Fourier series of a real periodic function has only P. cosine terms if it is even Q. sine terms if it is even R. cosine terms if it odd S. sine terms if it is odd Which of the above statement are correct? (A) P and S (C) Q and S (B) P and R (D) Q and R ECE - 2011 5. The trigonometric Fourier series of an even function does not have the (A) dc term (B) cosine terms (C) sine terms (D) odd harmonic terms

7.

Let g(t) = and h(t) is a filter matched to g(t). If g(t) is applied as input to h(t) , then the Fourier transform of the output is || (C) (A) ⁄

(B)

(D)

ECE- 2014 8. For a function g(t), it is given that t ∫ value if

t

y(t) = ∫

for any real t

(A) 0 (B) – 9.

∫ y t t s ⁄ (C) (D) ⁄

Consider a discrete time periodic signal x[n] = s

( ) Let

be the complex

Fourier series coefficients of x[n]. The coefficients { } are non-zero when k = Bm ± 1, where m is any integer. The value of B is ______. 10.

A Fourier transform pair is given by ( ) u

⇔

( )

Where u[n] denotes the unit sequence. The values of A is _______.

step

EE- 2006 1. x(t) is a real valued function of a real variable with period T. Its trigonometric Fourier Series expansion contains no terms of frequency = 2π (2k) / T;

th

th

th


GATE QUESTION BANK

k = 1, s s t rms r present. Then x(t) satisfies the equation (A) x(t) = x (t T) (B) x(t) = x (T t) = x ( t) (C) x(t) = x (T t) = x (t – T / 2) (D) x(t) = x (t T) = x(t – T / 2)

EE- 2008 4. Let x(t) = rect(t

x t

t≤

,

the Fourier Transform of x(t) + x( t) will be given by

5.

/

c s(

)

(C)

s (

)

(B)

c s(

)

(D)

s (

)

The frequency spectrum of a signal is shown in the figure. If this signal is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be |U 𝜔 |

(A) sinc( )

(C) 2sinc( )cos( )

(B) 2sinc ( )

(D) sinc( )sin( )

Let x (t) be a periodic signal with time period T. Let y(t) = x(t t0) + x(t + t0) for some t0. The Fourier Series coefficients of y (t) are denoted by . If = 0 for all odd k, then t0 can be equal to (A) ⁄ (C) ⁄ (B) ⁄ (D) 2T

EE- 2009 6. The Fourier Series coefficient, of a periodic signal x(t), expressed as ⁄ x(t) = ∑ are given by = j1; = 0.5 + j0.2; = j2; =0.5 – j0.2; = 2 + j1; and = 0; for | |> 2.Which of the following is true? (A) x(t) has finite energy because only finitely many coefficients are nonzero (B) x(t) has zero average value because it is periodic (C) The imaginary part of x(t) is constant (D) The real part of x(t) is even

1 KHz |U

and

/

(A)

(A)

x

zero otherwise). Then if sinc(x) =

x t { Which among the following gives the fundamental Fourier term of x(t)?

3.

)

(where rect (x) = 1) for

EE- 2007 2. A single x(t) is given by t≤

Signals and Systems

|

(B)

EE- 2010 7. x t is a positive rectangular pulse from t t t with unit height as shown in the figure. The value of | { where is the Fourier ∫ |

(C)

(D)

transform of x t } is 𝜔

th

th

th


GATE QUESTION BANK x(t)

1

1

t

(A) 2 (B) π 8.

(C) 4 (D) π

The second harmonic component of the periodic waveform given in the figure has an amplitude of

EE- 2014 11. For a periodic square wave, which one of the following statements is TRUE? (A) The Fourier series coefficients do not exist. (B) The Fourier series coefficients exist but the reconstruction converges at no point. (C) The Fourier series coefficients exist and the reconstruction converges at most points. (D) The Fourier series coefficients exist and the reconstruction converges at every point. 12.

+1

Let f(t) be a continuous time signal and let F( ) be its Fourier Transform defined by

0 T

T/2

Signals and Systems

∫ f t

t

t

Define g(t) by -1

t (A) 0 (B) 1 9.

(C) 2/π

The period of the signal πt

) is

(A) 0.4πs (B) 0.8πs

(C) 1.25s (D) 2.5s

EE- 2011 10. The Fourier series expansion ∑ f(t) = c s t s t of the periodic signal shown below will contain the following nonzero terms

13.

f(t)

0

(A) (B) (C) (D)

and , and , and and ,

u

u

What is the relationship between f(t) and g(t) ? (A) g(t) would always be proportional to f(t). (B) g(t) would be proportional to f(t) if f(t) is an even function. (C) g(t) would be proportional to f(t) only if f(t) is a sinusoidal function. (D) g(t) would never be proportional to f(t).

(D) √

x(t) = 8 s (

∫

A discrete system is represented by the difference equation [

]

+[

*

]

It has initial conditions . The pole locations of the system for a = 1, are (A) ± (C) ± (B) ± (D) ±

t

,

th

th

th


GATE QUESTION BANK

14.

A function f(t) is shown in the figure.

2.

If the waveform, shown in the following figure, corresponds to the second derivative of a given function f(t), then the Fourier transform of f(t) is

f t /

f t t

/ /

Signals and Systems

t

1

/ 1

The Fourier transform f f t is (A) real and even function of . (B) real and odd function of . (C) imaginary and odd function of . (D) imaginary and even function of . 15.

A signal is represented by |t| x t { |t| The Fourier transform of the convolved signal y t x t x t⁄ is

4.

The magnitude of fourier transform of a function x(t) is shown below in figure (a). The magnitude of fourier transform of another function y(t) is shown below in figure(b). The phases of and are zero for all . The magnitude frequency units are identical in both the figures. The function y(t) can be expressed in terms of x(t) as | | | |

A differentiable non constant even function x(t) has a derivative y(t), and their respective Fourier Transforms are X( ) and Y( ). Which of the following statements is TRUE? (A) X( ) and Y( ) are both real. (B) X( ) is real and Y( ) is imaginary. (C) X( ) and Y( ) are both imaginary. (D) X( ) is imaginary and Y( ) is real.

IN- 2006 1. The Fourier transform of a function g(t) is s

Then the function

g(t) is given as, (A) t xp( 3|t|) (B) c s t xp( 3t) (C) s t c s t (D) s t xp( 3t)

(D)

The Fourier series for a periodic signal is v s x t c s πt c s πt cos πt fu m t fr qu cy of the signal is (A) 0.2 Hz (C) 1.0 Hz (B) 0.6 Hz (D) 1.4 Hz

s

v

(C)

3.

s

16.

t

2

(A) 1 + sin (B) 1 + cos

s ( )s s ( )

+1

(A)

x( )

(C)

x t

(B)

x t

(D)

x( )

IN- 2007 5. Consider the discrete-time signal x[n] =( )

u

, where u

=,

≥

.

Define the signal y[n] as y[n] = x[ n],

th

th

th


GATE QUESTION BANK

∑ y (A) (B)

⁄

IN- 2011 9. Consider a periodic signal x(t) as shown below

qu s

(C) ⁄ (D) 3

⁄

Signals and Systems

x(t) 1

6.

Let the signal x(t) have the Fourier tr sf rm s rt s y(t) =

x t

t

where t

0 1

2 3

4 5 6

is an It has a Fourier series representation

arbitrary delay. The magnitude of the Fourier transform of y(t) is given by the expression | | | (A) | | (B) | | (C) .| |. (D) | |.| IN- 2008 7. The fourier transform x(t) = u t when u(t) is unit step function, (A) Exist for any real value of a (B) Does not exist for any real value of a (C) Exists if any real value of a is strictly negative (D) Exists if the real value of a is strictly positive

x t

/

∑

Which one of the following statements is TRUE? (A) for k odd integer and T = 3 (B) , for k even integer and T = 3 (C) for k even integer and T = 6 (D) , for k odd integer and T = 6 IN- 2014 10. x(k)is the Discrete Fourier Transform of a 6-point real sequence 𝑥(𝑛). If X(0)= 9 + j0, X(2)= 2 + j2, X(3)= 3 – j0, X(5)= 1 – j1,x(0) is (A) 3 (C) 15 (B) 9 (D) 18

IN- 2010 8. f(x), shown in the adjoining figure is represented by f x

∑{

The value of

c s x

s

x }

is f x

π

(A) 0 (B) π⁄

π

π

π

π

x

(C) π (D) π

th

th

th


t

GATE QUESTION BANK

Signals and Systems


6.

[Ans. C] c s

[Ans. A] x t x [ (t

)]

(

c s s c s c

[Ans. B] ( )

s

∫

f

| π

y

t

[Ans. A]

∫

⃡

t

(

t r ct ( )

{ t { t t {r c (

]

t } }

t )

r ct (

t

)}

1

|

½

-1

+1

3

s

) 7.

[Ans. D] The concept of matched filter assumes that the input signal is of the same form g(t) as the transmitted signal(except difference in amplitude ). This requires that the shape of the transmitted signal not change on reflection. h(t) = g( t ⇔ f (f) G f f f sr

± π π [Ans. A ] P & S are correct [Ans. C] S xp s f have sine terms.

)

t r ct ( )

)⃡

t

-3

5.

s c(

c

[

π

∫x t

4.

s

| π

+

c

) ys c

3.

*

Now r ct ( ) ⃡ s c(

∫ (

s

c s

)

Using scaling and shifting property. 2.

s

g(t) =

(fourier transform)

y t h(t)

g(t)

v

fu ct

g(t)

s ’t y(t)= h(t) * g(t) [convolution] th

th

th


GATE QUESTION BANK

8.

[Ans. B] Given

9.

x(t T) = x( t+T) but signal is periodic with period T. therefore x(t T) = x(t) therefore x(t) = x(T t Now since signal contains only odd harmonics i.e no terms of frequency π

t

S

y t

S

∫ y t t

∫

y i.e. no even harmonics. This means signal contains half wave symmetry this implies that x(t) = x(t / From (i) and (ii) x(t) = x(T t) = x(t T/2)

[Ans. *] Range 9.99 to 10.01 π x s * + /

P r S

fx π

Signals and Systems

/

π

2.

[Ans. A] The given signal x(t) is a periodic waveform with period T0 =2 T and satisfies half – wave symmetry:

So is non-zero for k = 1 and . Fourier co-efficient are periodic with N N= 10 So the value of k for which are

x(t) =

x (t ±

)

x t±

as shown

in Fig.1 Comparing We see m x 10.

±

x(t) 1

[Ans. *] Range 3.36 to 3.39 ( ) u

( )

( ) u ( ) ( )

s

u

( )

0 -1

↔ u u

Fig . 1 Fundamental frequency =

↔

x t

( ) ↔

EE 1.

m is any integer

x (t

)

shown

π/ = π/ in

Fig.

2

possesses Even symmetry and also half wave symmetry. Fundamental Fourier term of

( )

x t

( )

c st ( t)

Where M = peak to peak value of x t

[Ans. C] Since trigonometric fourier series has no sine terms and has only cosine terms therefore this will be an even signal i.e. it will satisfy x(t) =x( t) or we can write th

th

th


GATE QUESTION BANK x1(t)

4.

[Ans. C] ⁄ x t r ct t x t ≤t≤ = 0 otherwise

1 t

0

Signals and Systems

x t

1

∫x t

t

Fig. 2

∫

⁄ ) As x(t)= x (t Fundamental component of π x t c s [ (t )] π π π c s* t + π 3.

t

(

⁄

⁄

*

x t

≤t≤

∫x

t

t

(

)

(

M

) ⁄

[

⁄

⁄

0

1

f(in kHz)

x

Sampling interval , ms f z f f . Therefore Aliasing or overlap of the adjacent spectra in the sampled spectrum because f f . The sampled spectrum , U* f f ∑ f f as shown in Fig. 2. The resultant spectrum, U* is c st t f r ‘f’ s s w w c is the same figure given in option(b) |

x t

x

+

⁄

⁄

t ⁄ ⁄

[

⁄

s

s c(

⁄

⁄

]

c s ⁄

⁄

5.

⁄

⁄ s

)c s( ) π

[Ans. B] y t x t t x t t Since y(t) is periodic with period T. Therefore x t t x t t will also be periodic with period T.

|

fs M f

0 Fig. 3 |

|

|

2

3

is fourier series coefficient of signal x(t) therefore [ ] c s t s c f r

f |

f

-2

⁄

s

t

]

⁄

*

Fig. 1

⁄

+x

w r x t 0, otherwise

[Ans. B] Highest frequency of the input signal, fh = 1kHz as shown in its spectrum of Fig. 1 | | r| f |

)

-1 0 1 Fig. 2

f(in kHz)

th

th

th


GATE QUESTION BANK

i.e.

t

w

r t π

t 6. I.

Signals and Systems

∫[ ∫

± ± ± π/

]x t t

π ∫[ ∫

π

[Ans. C] A periodic signal x(t) has always infinite energy

III.

∫x t t

∫x t t

For a0 to be purely imaginary = 2j The first integral should be zero and the second integral should not be zero. i.e. xR(t) should be either 0 or an odd function. option (D) is not true. And xI(t) should be a constant or an even function so that the integral is not zero. option (C) is true. 7.

}

]x t t

t

t

option (A) is not true. The average value a0 of a periodic signal need not be zero. Infact a0 = j2 as given in this question option (B) is not true. The complex value of a0 indicates that x(t) should be complex. Let x(t) = xR(t) +jxI(t) Where xR(t) and xI(t) are the real and imaginary parts of x(t).

II.

{

π ∫x t x t t

π ∫ |x t | ∫ |x t |

π

π ∫

t

π ∫|

|

π

8.

[Ans. A] f(t) is odd  no cosine terms and no dc. Also, by inspection, half wave symmetry no sine even harmonics  no second harmonics amplitude = 0.

9.

[Ans. D] From the given signal x(t) π rad/sec π s c

10.

[Ans. D] Observe the given periodic f(t) shown in Fig. 1 f(t) A

[Ans. D]

t

0

∫|

|

∫ Fig. 1

∫[ ∫

t

(1) Its

t]

average ∫

value

,

f t t

(2) It has even symmetry. i.e; f(t) = f(-t) th

th

th


GATE QUESTION BANK

s t rms r s t f t (3) The function f1(t) = [f(t)-a0] has half – wave symmetry or odd harmonic symmetry: i.e., f t

f (

)

Signals and Systems

14.

[Ans. C] Given function is odd is imaginary and odd in

15.

[Ans. A]

Even harmonics are absent From observations (2) and (3) above, bn f r v u s f‘ ’ an f r v v u s f‘ ’ an f r

x t

x

s

x t

f1(t)

12.

u

u

f

t

16.

*

S

* *

|S P

s

s s

t

sr

(x t ) t

f ur r tr

sf rm

m

ry

IN 1.

[Ans. A] t Given ,

Use the F.T. pairs: t u t

*

s

| s

+

( )s

[Ans. B] x t v fu ct

y

+ putt s

s

t

+

*

( )

y

[Ans. A] Poles of the system are the roots of the | equation |S r

s

y t

f t ff t s v t is proportional to f(t) if f(t) is even. 13.

( )

s

s

[Ans. B] Since F( ) is FT of f(t) hence ∫

s

y

[Ans. C] Fourier series exists for periodic waves. Since it is a square wave, convergence occurs at most points

/ s

Convolution in time domain = multiplication of Fourier transforms t y t x t x( )

t

Fig. 2

11.

/ /

t x( )

0

s

s

+

If t replaced by ( t), (

+

is replaced by

u t

s

± th

th

th


GATE QUESTION BANK

u t

u

Signals and Systems

t So, y t

||

||

5. In general, K t 2.

t/

[Ans. C]

||

Given x(n) = ( ) u ||

t

y

[Ans. C] From the given plot, t

x

∑ y t

t

∑ ( )

∑ ( )

Take F.T on both sides Use the following pairs and properties: Let f(t)

rst t rm mm r t

t

f t t t

6.

[Ans. A] {x t t }

t

{

c s c s

4.

7.

[Ans. A] f z; f z f z u m t fr qu cy f x t f0=0.2 Hz (HCF of are the frequency) f f rm c f f harmonic

x t

|

t

|

}

| ||

|

[Ans. C] The fourier transform of u

t s

s >0 f

u t will exist

If a<0 8.

[Ans. A] The given function f(x) is periodic function with period, T = π F(x) is shown over one period from 0 to π in Fig.

[Ans. D]

|

t

|

c s 3.

u

|

|

The d.c value of f(x) = r We know that expansion in frequency domain result in compression in time domain and vice-versa. Here compression is done in frequency domain, So there will be expansion in time domain by same amount.

u

∫ f x x

rf x

As can be seen from Fig.

So for x ( ) th

th

th


GATE QUESTION BANK

Signals and Systems

f(x)

1.5 1.0 x

0 -1.0 -1.5

Fig.

9.

[Ans. B] Clearly, period of the signal x(t) is 3. So, T = 3 And

∫ x t t ∫x t t

so, ak=0 for k even integer. 10.

[Ans. A] Given 6 point DFT of x[n] with

y symm try pr p rt s

th

th

th


GATE QUESTION BANK

Signals and Systems

Z-Transform ECE - 2006 1. If the region of convergence of x [n] +x [n] is 1/3 < |z|<2/3, then the region of convergence of x [n] – x [n] includes

(A) < |z|< 3

(C) < |z|< 3

(B) < |z|< 3

(D) < |z|<

ECE - 2009 5. The ROC of Z-Transform of the discrete time sequence , x[n] = ( ) u

region of convergence of x[z] includes the unit circle. The value of x[0] is (A) 0.5 (C) 0.25 (B) 0 (D) 0.5 ECE - 2008 Statement for linked question 3 and 4: In the following network, the switch is closed at t = and the sampling starts from t = 0. The sampling frequency is 10Hz.

(D)

|z|

ECE - 2010 6. Consider the z-transform |z| z z z ; inverse z-transform x[n] is (A) 5 (B) 5 (C) 5 u u u (D) 5 u u u 7.

.

The

The transfer function of a discrete time LTI system is given by

z

Consider the following statements: S1: The system is stable and causal for ROC: |z|

x z x S mp r f z

is

(C) |z|

(B) |z|

. It is given that the

S

|z|

(A)

ECE - 2007 2. The z-transform x[z] of a sequence x[n] is given by x[z] =

( ) u

S2: The system is stable but not causal for z

tr

sf rm

ROC: |z| S3: The system is neither stable nor causal |z|

for ROC: 3.

The samples x [n] (n = 0, 1, 2, ...) are given by (A) 5 (1 – ) (C) 5 (1 – ) (B) 5 (D) 5

4.

The expression and the region of convergence of the z-transform of the sampled signal are

(A)

, |z| < e-5

(B) (C) (D)

, | z | < e-0.05 -

-

, | z | > e-0.05

Which one of the following statements is valid? (A) Both S1 and S2 are true (B) Both S2 and S3 are true (C) Both S1 and S3 are true (D) S1, S2 and S3 are all true ECE - 2011 8. Two systems z and z are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system z is

, |z| > e-5

th

th

th


GATE QUESTION BANK

z

x

(

=∑ (C) y [ n ] (D) y [ n ]

z

)

y

(A)

(C)

(B)

(D)

2.

(A)

| |

(C)

| |

(B)

| |

(D)

| |

of X(z) z (A) an-1 (B) an 3.

) u

(

+ z

z +1

) u

The region of convergence (ROC) of the ztransform of x[n] (C) s

(B) s |z|

(D) Does not exist

z

)H(z)

is

the

entire

Z-plane

(Except z = 0). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zero (C) two poles and one zero (D) two poles and two zeros

.

(A) s |z|

at z = a for n ≥ 0 will be (C) nan (D) nan-1

H (z) is a transfer function of a real system. (When a signal x[n] = (1 + j)n is the input to such a system, the output is zero. Further, the region of convergence (ROC) of

ECE - 2014 (

| |

with | |> a, the residue

Given X (z) =

( tx

z y (z)=∑ y (z)= z

EE - 2008

ECE/EE/IN - 2012 ⁄ | | ⁄ u 9. If x then the region of convergence (ROC) of its Ztransform in the Z-plane will be

10.

Signals and Systems

|z|

EE - 2014 11.

12.

Let x[𝑛] = 𝑥 𝑛]. Let z be the -transform of [𝑛]. If 0.5+j0.25 is a zero of z which one of the following must also be a zero of z (A) (B) / (C) / (D) The z-transform of the sequence x[n] is given by

z

with the region of

convergence|z|

. . Then, x [2] is_______.

EE - 2007 1. The discrete – time signal x[n]

X (Z) = ∑

4.

Let

causal signal x[n]. Then, the values of x[2] and x[3] are (A) 0 and 0 (C) 1 and 0 (B) 0 and 1 (D) 1 and 1 IN - 2008 1. The region of convergence of the ztransform of the discrete-time signal x[n] = 2nu[n] will be (A) | |>2 (C) | |> (B) | |< 2 (D) | |<

y[n]

Where denotes a transform – pair relationship, is orthogonal to the signal y (z)=∑

(B) y [ n ]

y (z)

be the Z-transform of a

IN - 2011 2. Consider the difference equation

z ,

(A) y [ n ]

z

y

x[n] = ( ) u

x

. Assuming the condition of

initial rest, th s ut

( ) z

(A) 3 ( ) th

th

and suppose that

f ry

≥

s

( ) th

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 252

GATE QUESTION BANK

(B)

( )

( )

(C)

( )

( )

(D)

( )

( )

4.

Signals and Systems

(A) | |<

(C) | |>

(B) | |<

(D) | |<

The transfer function of a digital system is given by:

IN - 2014 3. The system function of an LTI system is given by

; where

is real.

The transfer function is BIBO stable if the value of is: (A) 1.5 (C) 0.5 (B) 0.75 (D) 1.5

z z The above system can have stable inverse if the region of convergence of H(z)is defined as


r| [Ans. D] ROC {x x } {x x }(if are no pole-zero cancellation)

5. 2.

3.

[Ans. B] ROC is |z| x 0.5

u

z

[Ans. A] Recall the z – transform pairs: For the Right sided sequence: |z| | | u

x

For Left sided sequence: |z| u

[Ans. B] s

r |z|

z

z

there

|

z

.

/

( ) u

s

s

|z|

z z

( )

s

z

| |

z

|z|

For x(n) containing both the above terms the ROC is the overlapping region:

s

t Therefore samples

|z|

⁄

6. 4.

[Ans. A] For the given X(z) , note that the ROC is the entire z – plane except z z Recall the z – transform pair: Recall the time shifting property: If g(n) G(z) z z

[Ans. C] z

∑ ∑

z z

th

th

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 253

GATE QUESTION BANK

z For given X(z) , x(n)=5 7.

8.

[Ans. B] y x Taking z – transform of both sides z z z z z z For cascaded system z z z z z z z z z z z

9.

[Ans. C]

[Ans. C] Given : H(z)= Express H(z) in +ve powers of z and find the zeros and poles. z

z(

z)

(z

z

)

z (z (z

) (z

) )

Poles are at z = ¼ and z = ½ lying inside the unit circle in the z – plane. Consi r t t r p ss ’s:

Signals and Systems

| |

x

(i)|z| For a casual system h(n) =0, n< 0 and h(n) is a Right sided sequence and ROC is the region outside the circle passing through the right most pole z = ½ syst m s us For stability of LTI system unit circle should be inside the ROC syst m s st S sv st t m t

x

( )

( ) u

x

x

s y

∑ x

z | |

∑ x ( )

z

∑ x ( )

z

(ii)|z|

∑ x ( ) z

H(n) is a Left sided sequence and the system is not causal. t s t c us : f r ≥ As unit circle is not inside the ROC , the system is unstable. S s t – valid statement. (iii)

∑ x ( z) , z

( z) ,( z

|z|

)

)

( z

)

-

s

The ROC is the annular region between two circles bounded by the poles: At z

∑ ( z

| z|

| z

|

z |z|

H(n) is a two sided signal existing for sw sf r ≥ Syst m s not causal. As unit circle is not inside the ROC , the system is unstable. S s v st t m t t S S r tru S s t tru

|z|

x

| |

( ) u

x s| z

|

r|z| th

th

th


GATE QUESTION BANK

We draw the three ROC on the Z plane So, The over lapping region is given by

EE 1.

[Ans. B] Two discrete time signals x(n) and y (n) are said to be orthogonal, if ∑ x y whose * denotes complex conjugatio Or ∑ x y for real x(n) and y

|z| 10.

[Ans. C] v x (

) u

r(

) u

s qu

xt r r f c rc

fr

x

)

us

From the option given , according to the definition of Z.T:

|z|

y

[Ans. B] x x x z x z

12.

fx z s

| |

y

∑

z z y

is a zero of x(z)

x z

( ) u

y For

s s z r

z

z

c

us v r 11.

For the given x(n) , X(z) = ∑

|z|

) u ts

(

(

Signals and Systems

z

z

z

It can be seen that equation (2) is satisfied only when y(n) = y

x z

[Ans. *] Range 11.9 to 12.1 2.

x z

[Ans. D] For the given X(z) =

z (

)

Let Y(z) = z

taking Inverse z-transform using Frequency domain multiplication =time domain convolution x u u u u

z

Residue of Y(z)=

z z

z

with |z|>a

z |

z |

|

Note that the Question asked is to find x(n) given X(z) = with |z|>a , for which the

∑ ∑

answer is known through the following fundamental ZT pairs and properties. z |z| u z z z u z ( ) z z z |z|

∑ x S x

th

th

th


GATE QUESTION BANK

z z

(

)

|z|

| |

2.

|z|

z

| |

[Ans. A] (1

z

x

z z

z

r

put x

z ( ) u

z

z

(z

z

|z| )

z

z

z

(

[Ans. B] z

z z z z z

z z

z z z

x z x x

z

z

) (z

(

)

z

)

)

(z

)

z z

*

3.

( )

( ) +u

[Ans. C] For a system to be stable, it should contain Z=1 circle in its ROC r

∑ x u

(

)

z 4.

∑( ) z

) (z

)

} x

∑

(

z

y

{

) (z

z

z

z

[Ans. A] Given, x[n] = u x z

z z

(

z

1

z

IN 1.

|z|

y

z has ROC as entire Z-plane

which implies that it is polynomial. z s p 4.

|

[Ans. B] y

z

u 3.

| z

z | |

u

Signals and Systems

z

∑

[Ans. C] Only this option gives poles inside unit circle.

z

z

th

th

th


GATE QUESTION BANK

Signals and Systems

Laplace Transform ECE - 2006 1. Consider the function f(t) having Laplace transform F(s) = ; Re[s]>0. The final value of f(t) would be (A) 0 (C) (B) 1 (D)

1
<1

ECE - 2007 2. If the Laplace transform of a signal y(t) is Y(s)=

; then its final value is

(A) 1 (B) 0

ECE - 2011 6.

If

(C) 1 (D) Unbounded

ECE - 2009 3. An LTI system having transfer function and input x(t) = sin(t+1) is in steady state. The output is sampled at a rate rad/s to obtain the final output {y(k)}. Which of the following is true ? (A) s z r f r s mp frequency (B) s z r f r s mp frequency (C) s z r f r > 2 , but zero for < 2 (D) sz r f r > 2 , but nonzero for < 2 4.

Assuming zero initial conditions, the response y(t) of the above system for the input x t u t s v y (A) u t (B) u t (C) u t (D) u t

Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of s ∫ f (A) sF(s) – f(0)

(C) ∫

(B)

(D) [F(s) – f(0)]

s

ECE - 2010 5. A continuous described by y t y t t t

time y t

LTI

system x t t

s

f t

then the initial

and final values of f t are respectively (A) 0, 2 (C) 0, 2/7 (B) 2, 0 (D) 2/7, 0 ECE - 2013 7. A system is described by the differential equation

y t

x t .

Let x(t) be a rectangular pulse given by t x t , t rw s Assuming that y (0) =0 and at t =0, the Laplace transform of y(t) is (A)

(C)

(B)

(D)

EE – 2006 1. The running integrator, given by y t

∫x t

(A) has no finite singularities in its double sided Laplace Transform Y(s) (B) produces a bounded output for every casual bounded input (C) produces bounded output for every anticasual bounded input (D) has no finite zeros in its double sided Laplace Transform Y (s)

is x t

th

th

th


GATE QUESTION BANK

EE - 2010 Common Data for Questions 2 and 3 Given f t and t as shown below:

f(t) 1

(A) (B)

t

1

g(t) 1 0

3.

EE - 2012 5. Consider the differential equation y t y t y t t t t y wt y t | | t um r c v u

0

2.

Signals and Systems

t

5

3

t can be expressed as (A) t f t (B)

t

f(

(C)

t

f( t

(D)

t

f(

2 1

|

(C) 0 (D) 1

IN - 2006 1. Given, x t x t t xp the function x(t) is (A) xp t u t (B) xp t u t (C) t xp t u t (D) xp t u t

t u t

) IN - 2010 2. u(t) represents the unit step function. The Laplace transform of u(t- is

) )

The Laplace transform of

t is

(A) (B)

(A)

(C)

(B)

(D)

IN - 2013 3. The

(C)

discrete-time

transfer

function

is

(D) EE - 2011 4. Let the Laplace transform of a function f(t) which exists for t > 0 be s and the Laplace transform of its delayed version f t be s . Let (s) be the complex conjugate of s with the Laplace v r s t s s . If G(s) =

f

|

|

, then the inverse Laplace

transform of G(s) is (A) mpu s t (B) y mpu s t– (C) an ideal step function u(t) (D) an ideal delayed step function u(t–

(A) (B) (C) (D)

Non minimum phase and unstable. Minimum phase and unstable. Minimum phase and stable. Non-minimum phase and stable.

IN - 2014 4. The transfer function of a system is given by / s s The input to the system is ( )=sin100 t . In periodic steady state the output of the system s f u t y t s πt ). The phase angle ( ) in degree is ___________.

th

th

th


GATE QUESTION BANK

Signals and Systems


[Ans. C]

v

s

s} s sF(s) has poles ± , which lie on the imaginary axis. v u t r mc t pp For the given F(s) , f(t) = s t u t ≤f ≤ Hence the final value can be taken as : ≤f ≤ 2.

3.

s s t

r

[Ans. D] poles of S.Y(s) lie on right half of Splane) s = 1 is right s – plane pole Unbounded.

s

s

s

s tx t

s

s

s

u t

s

s

s

s s

s

s

s

s s Taking inverse Laplace transform on both side y t u t 6.

[Ans. B] s

s

f t

s

s

Initial value

[Ans. A] For the given LTI system

mf t

H(s) = mf t

The frequency response is given by

ms s

s

s (

)

s (

s s

s

)

t For the input x(t) = sin (t +1) with | | s (t y t ) t r /s c Y(.) is zero for all sampling frequencies sy t 4.

[Ans. B] 0∫ f

5.

Final value theorem mf t ms s ms

7. 1

[Ans. B] x t u t

s s

s s s

s

u t

x s

With zero initial condition.

s s Now taking Laplace transform

[Ans. B] y t y t x t y t x t t t t Taking Laplace transform on both side (assuming zero initial conditions). s s s s s s s s

s y s y s y s

th

th

sy s

y s

s s

s

s s

s

th

s


GATE QUESTION BANK

s s

t

s

Signals and Systems

f(

t

)

f(

t

)

Time shifting (delay by 2)

EE 1.

1

[Ans. B] y t

∫x t

0

t

∫

u t

∫

t

3.

t

5

[Ans. C] Form Fig. 3 in the above question, g(t) = u(t 3) u(t 5) Use L.T pair and property:

t

u t

s

u t *

4

Fig. 3

if we take x(t) as casual input ut x t u t y t

3

2

1

t

+

s f

t

s

s

s

s *

+

f r

t s

u

4.

[Ans. B] f t t

2.

[Ans. D] Time scaling (Expansion by 2) and Time shifting by 3 get g(t) from f(t) f(t) , f(t/2) and g(t) = f (

f t

) are shown

1 t

5.

s

s

s

s s |

| s | | s | ,

[Ans. D] +

Fig.1

Time scaling (expansion by 2)

y t

t

Converting to s – domain, s2y(s) – sy(0) – y (0) + 2[sy(s) – y(0)] + y(s) = 1 [s2 + 2s + 1] y (s) + 2s + 4 = 1

f(t/2) 1

0

→ s

| s

s

s G(s) t t ) The inverse LT of G(s) is a unit ideal impulse delayed by .

f(t)

1

t

s

fig. 1,2 and 3

0

→

y(s) = 1

2

find inverse lapalce transform y(t) [– 2 e–t te–t] u(t)

t

Fig. 2

= 2e–t + te–t – e–t | th

th

=2–1=1 th


GATE QUESTION BANK

IN 1.

Signals and Systems

[Ans. A] Given, x t x t t xp t u t taking Laplace transform on both sides we get, s

s

s

s Taking inverse Laplace transform of X(s) is x t xp t u t 2.

[Ans. C] Laplace transform of u(t) =1/s Use Time shifting property: If the L.T of f(t) is F(s), Then L.T of x(t) = f(t s s f u t-

3.

s

s

[Ans. D] Transfer function is Clearly all the zeros are on the RHS of imaginary axis so the system is nonminimum phase. Also, the poles z = 0.5 is inside the unity circle thus the system is stable.

4.

[Ans. *]Range 67 to 69 P s s t ( ) c s t

t π

th

th

th


GATE QUESTION BANK

Signals and Systems

Frequency Response of LTI systems and Diversified Topics ECE - 2006 1. A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t) where t

∑

A uniformly distributed random variable X with probability density function y 1

2.5

f x

If S(f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (A) S ≥S f (B) S f ≥ (C) S f s f (D) ∫ S f f

t

The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be (A) t (C) 0 (B) m t (D) m t t 2.

4.

x

2.5

(u x

u x

)

Where u(.) is the unit step function is passed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be (A) f y

(u y

(B) f y (C) f y

y

ECE - 2008 Statement for Linked Answer Questions 5 &6 The impulse response h(t) of a linear time-invariant continuous time system is given by h(t) = exp( 2t) u(t), where u(t) denotes the unit step function. 5. fr qu cy r sp s f t s system in terms of angular frequency is v y

u y

(B)

(D)

The output of this system to the sinusoidal input x (t) = 2cos (2t) for all time t, is (A) 0 (B) c s t π (C) c s t π (D) c s t π

7.

{x[n]} is a real-valued periodic sequence with a period N. x[n] and X[k] form Npoint Discrete Fourier Transform (DFT) pairs. The DFT Y[k] of the sequence

)

y

(D) f y

(C)

6.

y y

(A)

y

y /

y (u y

u y

)

ECE - 2007 3. If is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is NOT true? (A) |≤ (B) | (C) (D) The mean square value of the process is R(O)

y[n]= ∑

xrx

r is

(A) |X [k]|2 (B)

∑

r x

r

(C)

∑

x r x

r

(D) 0

th

th

th


GATE QUESTION BANK

8.

The probability density function (PDF) of a random variable X is as shown below.

0

1

0

(B)

(B) (C) (D)

1

x

The corresponding cumulative distribution function (CDF) has the form (A) CDF 1

1

(A)

PDF

1

ECE - 2009 10. The 4 –point discrete Fourier transform (DFT) of a discrete time sequence { 1, 0 ,2, 3} is (A) [ 0 , 2 +2j , 2 , 2 2j] (B) [2, 2 +2j , 6,2, 2j] (C) [ 6 , 1 3j , 2 , 1 + 3j] (D) [ 6 , 1 +3j , 0 , 1 3j] 11.

A system with transfer function H(z) has impulse response h(.) defined as h[2]=1 , h[3] = 1 and h[k] = 0 otherwise. Consider the following statements. S1: H(z) is a low-pass filter S2: H(z) is an FIR filter. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, and S2 is a reason for S1 (D) Both S1 and S2 are true, but S2 is not a reason for S1

12.

A white noise process X(t) with two-sided power spectral density / z is input to a filter whose magnitude squared response is shown below

1 x

CDF 1

0

1 (C)

1 x CDF

1

0

1 (D)

Signals and Systems

1 x

CDF 1

| X(t)

1

0

10kHz

1 x

f | Y(t)

10kHz

f

The power of the output process Y(t) is given by (A) (B) (C) (D)

1

9.

1

|x| |x| P x xp xp is the probability density function for the real random variable X, over the entire x axis. M and N are both positive real numbers. The Equation relating M and N is th

th

th


GATE QUESTION BANK

13.

If the power spectral density of stationary random process in a sinc-squared function of frequency, the shape of its autocorrelation is (A)

Signals and Systems

t

t

(B) t

t

(C)

(D)

ECE - 2010 14. The Nyquist sampling rate for the signal s t

is given by

(A) 400 Hz (B) 600 Hz 15.

16.

(C) 1200 Hz (D) 1400 Hz

For an N-point FFT algorithm with , which one of the following statements is TRUE? (A) It is not possible to construct a signal flow graph with both input and output in normal order (B) The number of butterflies in the m stage in N/m (C) In place computation requires storage of only 2N node data (D) Computation of a butterfly requires only one complex multiplication Consider the pulse shape s t as shown. The impulse response t of the filter matched to this pulse is

ECE - 2011 17. The first six points of the 8-point DFT of a real valued sequence are 5, 1 j3, 0, 3 j4, 0 and 3+j4. The last two points of the DFT are respectively (A) 0, 1 j3 (C) 1+j3, 5 (B) 0, 1+j3 (D) 1 j3, 5 ECE - 2013 18. The impulse response of a system is h(t)=tu(t). For an input u(t 1), the output is (A)

u t

(B)

u t

(C)

u t

(D)

u t

EC/EE/IN - 2013 19. For a periodic signal v(t)=30sin100t+10cos300t+6sin (500t π/ , the fundamental frequency in rad/s is (A) 100 (C) 500 (B) 300 (D) 1500

s t

ECE - 2014 20. Consider two real valued signals, x(t) band-limited to [ 500 Hz, 500 Hz] and y(t) band-limited to [ 1 kHz, 1 kHz]. For z(t) = x(t).y(t), the Nyquist sampling frequency (in kHz) is ______.

t

t

t

21.

t

An FIR system is described by the system function z

t

z

z

The system is th

th

th


GATE QUESTION BANK

(A) maximum phase (B) minimum phase (C) mixed phase (D) zero phase 22.

through the linear constant-coefficient differential equation y t y t y t x t t t Let another signal g(t) be defined as t t ∫ t t If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is________

Consider the periodic square wave in the figure shown. x

27. t

Let

z

z qz z z r z . The quantities p, q, r are real numbers. |r|

q

(A) also has a pole at

all . If (A) (B)

w m

r | (

)|

≤

for

, then b equals (C) / (D) /

25.

A modulated signal is y t m t c s πt where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t) is _____

26.

A causal LTI system has zero initial conditions and impulse response h(t). Its input x(t) and output y(t) are related

300

(B) has a constant phase response over the z-plane: arg| z | = constant (C) is stable only, if it is anticausal (D) Has a constant phase response over the unit circle: arg| | = constant

For an all-pass system z

≤

EE - 2006 1. A discrete real all pass system has a pole at z = 2 300 it, therefore

. If the

zero of H(z) lies on the unit circle then r = _____ 24.

∑x

Denote this relation as DFT(x). For , which one of the following sequences satisfies DFT (DFT (x)) = x ? (A) x = [1 2 3 4] (B) x = [1 2 3 2] (C) x = [1 3 2 2] (D) x = [1 2 2 3]

pz

Consider P

The N-point DFT X of a sequence x[n], ≤ ≤ is given by √

The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is closest in value to _______. 23.

Signals and Systems

EE - 2007 Statement for Linked Answer Questions 2 & 3: 2. A signal is processed by a causal filter with transfer function G(s). For a distortion free output signal waveform, G(s) must (A) Provide zero phase shift for all frequency (B) Provide constant phase shift for all frequency

th

th

th


GATE QUESTION BANK

(C) Provide linear phase shift that is proportional to frequency (D) Provide a phase shift that is inversely proportional to frequency 3.

4.

G(z) = z + z is a low-pass digital filter with a phase characteristics same as that of the above question if (1/3) (A) (C) / (B) (D) Consider the discrete-time system shown in the figure where the impulse response of G(z) is g[0] = 0, g[1] = g[2] = 1, + +

Σ

G(z)

K

The system is stable for range of values of K (A) [ 1, ½] (C) [ 1/2, 1] (B) [ 1, 1] (D) [ 1/2, 2] EE - 2008 5. A signal x (t) = sinc( t) where constant (s c x

filter with a cut-off frequency of 20 Hz. The resultant system of filters will function as (A) An all-pass filter (B) An all-stop filter (C) A band stop (band-reject) filter (D) A band-pass filter EE - 2014 7. A sinusoid x(t) of unknown frequency is sampled by an impulse train of period 20 ms. The resulting sample train is next applied to an ideal lowpass filter with a cutoff at 25 Hz. The filter output is seen to be a sinusoid of frequency 20 Hz. This means that x(t) has a frequency of (A) 10 Hz (C) 30 Hz (B) 60 Hz (D) 90 Hz IN - 2006 1. A digital filter has the transfer function H(z) =

If this filter has to reject a

50 Hz interference from the sampling frequency signal should be (A) 50 Hz (C) (B) 100 Hz (D)

is a real

) is the input

to a Linear time invariant system whose impulse response h(t) = sinc ( t) where s real constant. If min ( t s the minimum of and and similarly max ( ) denotes the maximum of and and K is a constant, which one of the following statements is true about the output of the system? (A) It will be of the form K sinc( t) where = min ( ) (B) It will be of the form K sinc( t) where = max ( ) (C) It will be of the form K sinc( t) (D) It cannot be a sinc type of signal

Signals and Systems

2.

the input, then for the input 150 Hz 200 Hz

The spectrum of a band limited signal after sampling is shown below. The value of the sampling interval is

f(Hz) 100 0

100 150

350 400

(A) 1 ms (B) 4 ms 3.

600

(C) 2 ms (D) 8ms

A digital measuring instrument employs a sampling rate of 100 samples/second. The sampled input x(n) is averaged using the difference equation y x x x x /

EE - 2011 6. A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high-pass th

th

th


GATE QUESTION BANK

For a step input, the maximum time taken for the output to reach the final value after the input transition is (A) 20ms (C) 80ms (B) 40ms (D) IN - 2007 4. Let x(t) be a continuous-time, real-valued signal band-limited to F Hz. The Nyquist sampling rate, in Hz, for y(t) = x (0.5t) + x(t) x(2t) is. (A) F (C) 4F (B) 2F (D) 8F

IN - 2010 8. 4-point DFT of a real discrete-time signal x[n] of length 4 is given by X[k ], n= 0,1 2, 3, and k =0, 1, 2 ,3.It is given that X[0] =5,X[1]=1+ j1, X[2]=0.5. X[3] and x[0] respectively are (A) 1 j, 1.875 (B) 1 j, 1.500 (C) 1+j, 1.875 (D) 0.1 j0.1, 1.500 9.

A digital filter having a transfer function H(z) =

y[n 1].

If y[n] = 0 for n < 0 and x[n] = [n], then y [n] can be expressed in terms of the unit step u[n] as (C) u (A) ( ) u (D) u (B) ( ) u

10.

IN - 2009 6. An analog signal is sampled at 9kHz. The sequence so obtained is filtered by an FIR filter with transfer function H[z] z . One of the analog frequencies for which the magnitude response of the filter is zero is (A) 0.75kHz (C) 1.5kHz (B) 1kHz (D) 2kHz 7.

The transfer function H(z) of a fourth order linear phase FIR system is given by z z z z . Then G(z) is (A) z z (B)

z

(C)

z

(D)

z

z z z

is implemented

using Direct Form – I and Direct Form – II realizations of IIR structure. The number of delay units required in Direct Form – I and Direct Form – II realizations are, respectively (A) 6 and 6 (C) 3 and 3 (B) 6 and 3 (D) 3 and 2

IN - 2008 5. Consider a discrete-time system for which the input x[n] and the output y[n] are related as y[n] = x[n]

Signals and Systems

H(z) is a discrete rational transfer function. To ensure that both H(z) and its inverse are stable its (A) poles must be inside the unit circle and zeros must be outside the unit circle (B) poles and zeros must be inside the unit circle (C) poles and zeros must be outside the unit circle (D) poles must be outside the unit circle and zeros should be inside the unit circle

IN - 2011 11. The continuous time signal x(t) = c s πt s πt is sampled at the rate 100 Hz to get the ∑ signal x t x t , = sampling period The signal x t is passed through an ideal low pass filter with cutoff frequency 100 Hz. The output of the filter is proportional to th

th

th


GATE QUESTION BANK

(A) (B) (C) (D) 12.

c s c s c s s

πt πt πt πt

s s

Which one of the following statements is TRUE? (A) This is a low pass filter (B) This is a high pass filter (C) This is an IIR filter (D) This is an FIR filter

πt πt

Shown below is the pole-zero plot of a digital filter Im Z plane

Re

6th order pole at the origin

Signals and Systems

IN - 2014 13. A discrete-time signal x[𝑛] is obtained by sampling an analog signal at 10 kHz. The signal x is filtered by a system with impulse response ℎ[𝑛] = 0.5{ [𝑛]+ [𝑛 1]}. The 3dB cutoff frequency of the filter is: (A) 1.25 kHz (C) 4 .00 kHz (B) 2.50 kHz (D) 5.00 kHz


[Ans. B] m(f)

500 500

20kHz

f

= m(t) in time domain.

500

2.

m(t)*g(t)=m(F) G(f) G(f)

20kHz

500

[Ans. B] Sample

of

random

variable

After transformation smaple space of random variable Y=(0, 1) Hence f y y y pt ‘ ’ s t sf s

f

m f

space

f

3.

[Ans. C] Autocorrelation function is an even function

4.

[Ans. B] PSD is always a positive quantity

500 500 20kHz 20kH z pass filtering with f After low z

th

th

th


GATE QUESTION BANK

5.

[Ans. C]

9.

[Ans. A] P P x

u t H(s) =

h(t) =

Signals and Systems

|x|

xp

xp

|x|

∫P x x 6.

[Ans. D] Input x(t) = 2cos(2t) Frequency response π

| |

r∫

Output

} x

x

r π

r

π

π 10.

x t t π

π

| |

r ∫{

x

t

{

[Ans. D] 4-point DFT of sequence {1,0,2,3} is given as [

}

][ ]

{ }

[

π

]

[

]

π 11. c s t √

√

s

t

[ c s t √ c s t

[Ans. A]

8.

[Ans. A] P f x

x

s

t]

h[n] is high pass filter & FIR filter.

π

c s t 7.

√

12. π

t t

u t u t

[Ans. A] h[2] =1 and h[3]= 1 } h [n] = {

[Ans. B] PSD of white noise / z PSD of output | f | f f | f | utput s p w r

tu t

∫

∫f x x

f f r

Integral of increasing ramp signal is increasing parabola and integral of decreasing ramp signal is decreasing parabola.

(

r|

u

f | curv

)

f th

th

th


GATE QUESTION BANK

17.

13.

[Ans. B] PS 18.

Signals and Systems

[Ans. B] For 8-points DFT; x x x x x x it is conjugate symmetric about x[4] x x [Ans. C]

= FT

t

tu t s

s

c s

s t

[c⏟ s

s

*

πt

c s ⏟

πt ]

[Ans. A] t

[Ans. D] (A), (B) and (C) are wrong

s

t

c s

t π⁄ ]

are rational

20.

[Ans. C] Impulse response h(t) of the filter matched to the input pulse s(t) existing from t =0 to t= T is given by h(t) =s[-(t-T)] the operation of time reversal and then time delay of T should be performed on s(t) , to get h(t). This is equivalent to the following statement; scan the sketch of s(t) backward from t =T to t =0 and draw the sketch of h(t) forward from t=0 to t =T. s(t) & h(t) are shown below in Fig.

21.

fu

m

t

p r

fu

m

t fr qu

=

cy

p r

[Ans. *] Range 2.99 to 3.01 f z z f f z

z

[Ans. C] z

z z

z z

z z

z

z

z

z

z z

So, zeros are at

T

ut

+

s [

Maximum frequency in s t is 600Hz ( s t ) SO, Nyquist sampling rate 2 + z

16.

s

c s 19.

15.

s

y t

s

s

⁄s

y s [Ans. C]

u t

⁄p

s c fu ct 14.

and



T

th

and

.

When all the zeroes are inside the unit circle minimum phase th

th


GATE QUESTION BANK



When all the zeros are outside the unit circle Maximum phase When some zeros are inside unit circle some outside unit circle mixed phase



22.

[Ans. *] Range 0.50 to 0.52 Consider one period from t t s c x t t t t

| S | 23.

Pz

(P

z z

z

qz

z

z rz

z z

r

f

z sz

r 25.

[Ans. *] Range 9.5 to 10.5 u t s y t m t c s πt to recover m(t) minimum required rate should be twice the minimum frequency of m(t) f f z

26.

[Ans. *] Range 0.99 to 1.01 y y y t x t t t y y * y t + t t s s s s s t t ∫ t t

)

z s

rz (z

) (z

z*

z )

r z

(z

) (z

s *

+

r Since, this zero lies on unit circle |z| z z S c

v

|r|

27.

r r

s

t

s s

s

+ s s

[Ans. B] x

r

r

s s s s

x t

s s s s So number of poles G(s) = 1

)

One zero is at origin. The other zero lies at z

z w

p

z

z

z

Poles of H(z) is z r p ss syst m w

)

(q

z

z

r

[Ans. *] Range z

z

z

| |

z

[Ans. B]

t

c s π c s π π m rm c | | c s π | | π

P w r f

24.

Signals and Systems

√

[

]

r √

th

[

th

]

th

x


GATE QUESTION BANK

[

] x

[

z

z

] x

r

z z

z

z z S t syst m w st f t’s ut r most pole will lie inside the unit circle. Location of poles

[Ans. C] For causal system, if all poles are inside the unit circle then system is stable, and converse is true for anti-causal system. 2

z z

r

Now check for each Options EE 1.

Signals and Systems

±√

z √

2

√ √

If anti-causal, ROC is |z| Syst m s st . 2.

3.

4.

[Ans. C] For distortion free output phase shift must be linear function of frequency i.e. proportional to frequency this is because delay to all frequency component will be equal.

sc

t

s s t sf

y pt

(A) 5.

[Ans. A] x(t) = sinc( t h(t) =s t x(t) = s π t

[Ans. A] For distortion free output phase shift must be linear function of as will as all the frequency component must be amplified by same amount so z corresponds to frequency . While z corresponds to frequency 3 . In order to have same amplification of frequency component at ,

s π t s π t 1/

𝜔

π

π

h(t)= s π t 1/

𝜔

[Ans. A] Given g(1) =g(2) =1. Otherwise 0 i.e. g[n] = therefore G(z) = z z Therefore overall transfer function of closed loop system z z z

π

π

y(t) = h(t) * x(t) 1/

if y π

th

th

π

th


GATE QUESTION BANK 1/

or

[Ans. C] Sinusoid x(t) frequency say f .Hz s mp

Or Y(j𝜔

So

7.

if

π

y(t) = s π t c

t

f

y(t) = s

c

t

f

z utput z u t frqu IN 1.

[Ans. D] The frequency response, f of LPF with cutoff frequency, 30 Hz is shown in Fig. 1. The frequency response f of HPF with cutoff frequency, 20 Hz is shown in Fig. 2 If these two filters are cascaded, the overall frequency response of the resultant system H(f) = H1(f) H2(f), shown in Fig. 3, represents a BPF with Bandwidth, B = 10 Hz H1(f)

0 10 20 30 40

2.

0

f

f

f

f

ms=4ms

[Ans. B] Since output y depends on input, such as no delay, delay by 1 unit, delay by 2 unit, delay by 4 unit, so it will sum all the sample after 4Ts (maximum delay), to get one sample of y[n].

f

t 4.

0

20

30

s t

Where fs is the sampling frequency i.e., X(f) is repeated at ±f ± f ± f tc From the given spectrum of Xs(f) in the question, fs = 250 s/sec S mp t rv

Fig. 2 H(f)

20

f ∑

Ts=

20 30

cy tr

[Ans. B] Let X(f) be the spectrum of the bandlimited signal, x(t). let Xs(f) be the spectrum of the sampled signal xs(t)

3.

20

z

[Ans. A] In order to reject the 50 Hz interference, sampling frequency must be as low as the noise frequency (so that they are separated in time domain and has less interference) So, f z

Fig. 1 H2(f)

30

cy

f

f

y(t) = s

30

fr qu

π

So output is of the form k sinc( t Where m 6.

Signals and Systems

ms c

[Ans. C] x(t) band limited to FHz i.e., fm = FHz y(t) = x(0.5t) + x(t) –x(2t) is x(t) – maximum frequency =

f

z

Fig. 3 th

th

th


GATE QUESTION BANK

x(0.5t) – maximum frequency =

Signals and Systems

z

z

z z

x(2t) – maximum frequency = 2FHz. Y(t) – maximum frequency

z z

= max (F, Nyquist rate = =2 5.

S

m x frequency z

we know that for linear phase filter

[Ans. A]

for ≤ ≤ where M is the number of points filter here, Now,

Given y(n) = x(n) – y y y

y

x

Take z – transform: z [

z

]

z

H(z) = For the given x(n) = z

z

So, : : : : This is only in option (a).

Use the standard Z.T. pair: z |z| | | u z y

(

8.

[Ans. A] As signal is real, X[3] = 1 – J and

|z|

) u

∑ 6.

[Ans. B] z ( )

z c s

| (

| (

)|

√

7.

c s

s

9.

[Ans. B]

√

c s

Given H(z) =

√

s

This T.F corresponds to a 3rd order digital filter D F – I realization requires 2N delays D F – II realization or canonic realization with minimum number of delay elements requires N delays. Where N is the order of the order of the filter – I requires 6 delays D F – II requires 3 delays

)|

so, | ( )| w r π f

s

s

≤ , if

≤ π π

π f r

z

[Ans. A] z Let, z then,

z

z z

z z

th

th

th


GATE QUESTION BANK

10.

11.

12.

[Ans. B] The discrete time system with rational transfer function , H(z) is stable if the poles of H(z) lie inside the unit circle in the z – plane. r z ts v rs / z t stable , both the poles and zeros of H(z) must lie inside the unit circle in the z – plane. [Ans. B] Highest frequency component in x(t) is 150 Hz. So, the Nyquist sampling rate is 300 Hz. But, x(t) is sampled at 100 Hz. While cos(100 πt) with frequency 50 Hz will be recovered satisfactorily after passing through the low – pass and sin(300πt) will get aliased resulting in filter output sin (100 πt). Cos (100 πt s ’t c tribute to aliasing.

z

z

)(

[z

⌊z

z

[(z

[ z

(z z

) z

]

z

z

z

⌋

z

z

z

As the impulse response, h(n) = Inverse Z.T of H(z) has only finite duration =7 samples, the given digital filter is an FIR filter. 13.

[Ans. B] Given that ( )

[

] c s

| (

√

)| √

cut ff

z

πr π

]

th

th

√

s

c s( ) |

| π

√ ⁄s mp r

s

c s

c s

√

s

c s

c s

)

) z

z

z

c s z

z ]

[Ans. D] From the given pole – zero plot of the digital filter, the system function (

Signals and Systems

√ π

r z π

⁄s mp

π

th

z


GATE QUESTION BANK

Control Systems

Basics of Control System ECE - 2008 1. A signal flow graph of a system is given below

ECE - 2010 2. The transfer function Y(s)/R(s) of the system shown is (s)

-

(s) s

1 1

1/s

u s

1/s 1

u

1

1/s

(A) 0

(C)

(B) The set of equations that correspond to this signal flow graph is x x (A) (x ) = [ ] (x ) + x x [

[

[

[

s

=

[

x ] (x ) x

+

=

[

=

n

t

tr ns r un t on

( )

+ 4.

[

x ] (x ) x

(s)

w

( ) x x ]( ) x

(s)

s

(s)

u ] .u / x [x ] x

(D)

(s)

u ] .u / x x ( ) x

(C)

ECE - 2014 3. For the following system,

u ] .u / x (x ) x

(B)

(D)

s

( ) ( )

s

s s(s s ( ) s(s ( )

s s

) )

Consider the following block diagram in the figure. (s)

(s)

+

u ] .u / (A)

(s) s (s) (C)

(B)

(D)

tr ns r un t on

th

th

th

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30 Cross, 10 Main, Jayanagar 4 Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 276

GATE QUESTION BANK

EE - 2007 1. The system shown in figure below. c0

b0

b1

Σ 1/s

Σ

EE - 2010 3. As shown in the figure, a negative feedback system has an amplifier of gain 100 with 10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately:

C1 Σ

Σ

Σ

Σ1/s

Control Systems

p

a

a

1

0

Can be reduced to the form X

+ Σ +

X

Y

Y

Σ

P

⁄

P

(A) 10 1% (B) 10 2%

(C) 10 5% (D) 10 10%

Z

With (A) X = s

EE - 2014 4. The closed-loop transfer function of a

,

Y= ⁄(s

(B) X = 1, Y =

) ) ⁄(s

s ( s

(C) X =

(D) X

s ⁄(s

) ⁄(s

s

The steady

)

)

s

The signal flow graph of a system is shown below. (s) is the input and (s)is the output.

,

( s

(

state error due to unit step input is________ 5.

s s

(s)

system is

s

) s

,

s

(s)

s

)

EE - 2008 2. A function y(t) satisfies the following differential equation,

()

(s)

s

Assuming n the input-output transfer

+ y(t) = (t),

( )

(s)

where (t) is the delta function. Assuming zero initial condition and denoting the unit step function by u(t), y(t) can be of the form (A) (C) u(t) (B) (D) u(t)

( )

( )

(s)

( )

(s)

( )

(s)

( )

th

(s)

th

, function

of the system is given by s s

s s

s

s s

s

s s

s

s

th


GATE QUESTION BANK

6.

The block diagram of a system is shown in the figure (s)

(s)

s

(s)

IN - 2009 3. A filter is represented by the signal flow graph shown in the figure. Its input x(t) and output is y(t). The transfer function of the filter is

If the desired transfer function of the system is ( )

Control Systems

1

1

X(s)

k

then G(s) is

( )

(A) 1 (B) s

Y(s)

(C) 1/s (D)

1/s

1/s

1

(A)

1 s

(B)

s 1 s(s2  2)

s(s2  1) s2  2

(A) (B)

(D)

(C)

K

(

(C)

)

(D)

(

) )

1

1

1

C

The transfer function (C/R) of the system is

IN - 2007 2. A feedback control system with high K, is shown in the figure below: R(s) +

)

Y(s)

1

(C) 2

1

(

1

IN - 2011 4. The signal flow graph of a system is given below. R

1/s

(

(B)

s

1

1

(A)

IN - 2006 1. The signal flow graph representation of a control system is shown below. The Y(s) transfer function is computed as R(s)

R(s)

1/S

k

G(s)

C(s)

(D)

)

(

) (

)

(

) (

)

(

) (

)

(

)

IN - 2012 5. The transfer function of a Zero-orderHold system with sampling interval T is

H(s)

Then the closed loop transfer function is. (A) sensitive to perturbations in G(s) and H(s) (B) sensitive to perturbations in G(s) and but not to perturbations H(s) (C) sensitive to perturbations in H(s) and but not to perturbations G(s) (D) insensitive to perturbations in G(s) and H(s)

(

(A)

(

)

(C)

(B)

(

)

(D)

IN - 2013 6. The complex function tanh(s) is analytic over a region of the imaginary axis of the complex s- plane if the following is TRUE everywhere in the region for all integers n (A) Re(s) =0 (C) Im(s) ≠ (

(B) Im (s) ≠ n

th

th

(D) Im(s)≠

th

)


GATE QUESTION BANK

7.

The signal flow graph for a system is given below. The transfer function

( ) ( )

for

this system is given as

Control Systems

IN - 2014 8. A plant has an open-loop transfer function, (s)

(s)

s

s

(A)

(C)

(B)

(D)

(s )(s )(s ) The approximate model obtained by retaining only one of the above poles, which is closest to the frequency response of the original transfer function at low frequency is

(s)

(A)

(C)

(B)

(D)

Answer Keys and Explanations ECE 1. [Ans. D]

u

2.

[Ans. B] (s)

(s) x

(s)

s

x s x

u

(s)

x x

(s)

(s) [

s (s) (s)

o (s) x x x u x x x u x x x x x u x x [ ] [ ][ ] [ ] 0u 1 x x One can denote any state by any name So, that answer is x x u [x ] = [ ] (x ) + [ ] .u / x x

(s)

s

r or 3.

]

s

(s) (s) s

(s) (s)

s

[Ans. D] When x (s) (s)

(s) s s s

y(s) x (s)

th

th

s s(s

th

)


GATE QUESTION BANK

4.

[Ans. C]

Control Systems

These are four individual loops

(s)

(s)

s

(s)

(s)

s

s

s

s

s s s

s s All the loops touch forward paths (

(s)

(s)

(s)

(s)

s s s s s ng M son’s g n ormul (s) (s) s

(s) (s) (s) (s)

(s) (s) EE 1.

)

s

(

)s ( s) (s s) (s

( (

) s

) s

)

()

(s) Σ

[Ans. D] Σ

(s)

s

Σ

Σ

(s) xy (s) yz Comparing eg. (I) and (ii), we get s xy s s s yz s s Hence option (D) is correct

s s

Signal flow graph of the block – diagram

s ( )

2.

s

[Ans. D] Take LT on both sides,

( )

s.Y(s)+Y(s) =1  Y(s) = (  y(t) =

There are forward paths 3.

)

u(t)

[Ans. A] T.F =

th

th

=10

th


GATE QUESTION BANK

6.

Also,

Control Systems

[Ans. B]

 overall change in T.F for G = 10 is

(s)

(s)

s

= 1%

(s)

1% 4.

[Ans. 0] (s)

s

(s)

s ( or un t st p)

s

(s)

s(s

y(t)

y st t

s(s

s

1

)

( ) ( ) ( ) ( ) So put G(s) = s ( ) ( )

( )

IN 1.

Loops: (

s

(s) (s) n

M

( ) ( ( )

)

( )

s

s

M

s

s

s

) s

∑

num

( )1

[Ans. A]

)

(

s

( )

( )

0

s

( ) s

1 C(s)

s

1

rror

s

G(s)

1/s

s

[Ans. C] Using Mason’s gain formula Paths:

s

1

R(s)

)

s

l m sy(s)

t 5.

Convert the given block diagram to signal low gr p n us M son’s rul 1

There individual loops with gains: ro p t s

( )( )( s s

) ( )( )( s s

0

1

∑ (

)

s

(s) (s)

s

M

)

s

(

M

)( )( s

.

/

.

/

s (s) (s)

( .

/

s

s

)

2. (

[Ans. C] In this case, open loop gain sensitivity is measured with respect to KG. Because here open loop gain = KG

)

s s s

(

s th

th

(

)

)

(

th

)(

)


)

s

GATE QUESTION BANK

)( ) (

(

s ount)

6.

Control Systems

[Ans. D] t n (s)

(

)(

(

)

( or l rg s

(

s

)

)

For analytical region there should be no poles ≠ os sn os sn ≠ os ≠ ( n ) ≠

)

(

(

)

s n (s) os (s)

)

So close loop transfer function is sensitive to perturbation in H(s) but not to perturbation in G(s) for large k values.

7.

[Ans. A] u(s)

3.

s

y(s)

forward path =

. /; (

( )

=

( )

4.

s

[Ans. A]

(

)

( (

)

|

) )

Loop =

[Ans. C]

(s) =1 (

)

(

) 8. (

( 5.

[Ans. A] Writing the given open loop transfer in

) )

form of

[Ans. A] The transfer function of a Zero-orderHold system having a sampling interval T is

(

g t

.

/.

/.

/

So, at low frequencies, the approximate model equivalent to original transfer function will

)

(

th

)

th

th


GATE QUESTION BANK

Control Systems

Time Domain Analysis ECE - 2006 1. In the system shown below, x(t) = (sin t) u(t). In steady-state, the response y(t) will be

x(t) (A) (B) 2.

3.

s √ √

y(t)

sin .t

/

(C)

sin .t

/

(D) sin t –cos t

√

ECE - 2008 6. Step responses of a set of three secondorders underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of three systems? (A) j

sin t

The unit impulse response of a system is (t) t For this system, the steady-state value of the output for unit step input is equal to (A) 1 (C) 1 (B) 0 (D)

(B) j

The unit-step response of a system starting from rest is given by (t) – or t . The transfer function of the system is (A)

(C)

(B)

(D)

(C) j

ECE - 2007 4. The frequency response of a linear, timeinvariant system is given by H(f)=

5.

(D)

. The step response of the system is (A) 5 (1 )u(t) (C) (1

)u(t)

(B) 5 (1

)u(t)

)u(t)

(D)

(1

j

The transfer function of a plant is (s)

(

)(

)

.

The

second-order

approximation of T(s) using dominant pole concept is (A) (B)

( (

)( )(

) )

(C) (D)

( (

7.

A linear, time-invariant, causal continuous time system has a rational transfer function with simple poles at s and s , and one simple zero at s . A unit step u(t) is applied at the input of the system. At steady state,

) )

th

th

th


GATE QUESTION BANK

the output has constant value of 1. The impulse response of this system is (A) , xp ( t) xp ( t)- u(t) (B) , xp ( t) xp ( t) xp ( t)- u(t) (C) , xp ( t) xp ( t)- u(t) (D) , xp ( t) xp ( t)- u(t) 8.

Group I lists a set of four transfer functions. Group II gives a list of possible step responses y(t). Match the step responses with the corresponding transfer functions Group I

Control Systems

4. y(t) 1

t

(A) (B) (C) (D) 9.


The magnitude of frequency response of an underdamped second order system is 5 at 0 rad/sec and peaks to

s s

s

s

s

s Group II 1.

√

at

√ rad/sec. The transfer function of the system is,

s

(A)

(C)

(B)

(D)

ECE - 2009 10. The unit step response of an underdamped second order system has steady state value of 2. Which one of the following transfer function has these properties?

y(t)

t

2. y(t)

(A)

(C)

(B)

(D)

ECE - 2010 11. A system with the transfer function

1

( )

has an output

( )

t

3. y(t)

1 t

y(t)

os . t

x(t)

p os . t

/ for the input signal /.Then, the system

p r m t r ‘p’ s (A) √

(C) 1

(B)

(D)

√

√

ECE - 2011 12. The differential equation y

x(t)

describes

a

system with an input x(t) and an output th

th

th


GATE QUESTION BANK

y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform

Control Systems

r

(s)

y

(s)

Which one of the following compensators C(s) achieves this? (C) (s ) (A) . /

(A) y(t)

(B) t

(D)

/

.

/

15.

The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is ________.

16.

The input u(t) where u(t) is the unit step function, is applied to a system

(B) y(t)

t

.

(C)

with transfer function

y(t)

. If the initial

v lu o t output s t n t v lu of the output at steady state is _______. t

17.

(D) y(t)

The steady state error of the system shown in the figure for a unit step input is _________ (s) r(t)

(s)

(s) (t)

s

(t)

t s

ECE - 2012 13. A system with transfer function ) (s )(s (s) (s )(s )(s ) is excited y s n( t). The steady-state output of the system is zero at (A) r s (C) r s (B) r s (D) r s ECE - 2014 14. For the (s)

(

following )(

)

feedback

18.

(s)

(s) s(s

(A) 16 (B) 4

system

. The 2%-settling time of

the step response is required to be less than 2 seconds.

For the second order closed-loop system shown in the figure, the natural frequency (in rad/s) is )

(C) 2 (D) 1

EE - 2007 1. Consider the feedback control system shown below which is subjected to a unit step input. The system is stable and has th

th

th


GATE QUESTION BANK

the following parameters kp = 4, ki = 10, n ξ 1

0

The transfer function of a system is given as

z

s +

5.

Σ

KP

s

ξ s

+

The steady state value of z is (A) 1 (C) 0.1 (B) 0.25 (D) 0

(

(A) (B) (C) (D)

+ Σ

Control Systems

)

. The system is ___________

An over damped system An underdamped system A critically damped system An unstable system

EE - 2009 6. The unit-step response feedback system with

of a unity open loop

transfer function G(s) =

Statement for Linked Answer Question Q.2 & Q.3 R-L-C circuit shown in figure,

(

)(

)

is

shown in the figure. The value of K is 1

m

0.75 0.5 0.25 0 0

2.

3.

For a step- input overshoot in the output will be (A) 0,Since the system is not under damped (B) 5% (C) 16% (D) 48% If the above step response is to be observed on the non-storage CRO, then it would be best have the as a (A) Step function (B) Square wave of frequency 50 Hz (C) Square wave of frequency 300 Hz (D) Square wave of frequency 2.0KHz

EE - 2008 4. The transfer function of a linear time invariant system is given as G(s) =

1

2 Time

(A) 0.5 (B) 2

3

4

(C) 4 (D) 6

EE - 2010 7.

For the system

(

)

, the approximate

time taken for a step response to reach 98% of its final value is (A) 1s (C) 4s (B) 2s (D) 8s EE - 2011 8. The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is r(t) 10

. The steady state value of

the output of the system for a unit impulse input applied at time instant t=1 will be (A) 0 (C) 1 (B) 0.5 (D) 2

t

1s

(A) 0 (B) 0.1 th

th

(C) 1 (D) 10 th


GATE QUESTION BANK

IN - 2007 Common Data for Questions 1 & 2 The following figure represents a proportional control scheme of a first order system with transportation lag. R(s)

(s

1.

2.

at which the response to a unit step input reaches its peak is seconds 4.

The damping coefficient for the closed loop system is (A) 0.4 (C) 0.8 (B) 0.6 (D) 1

5.

The value of p is (A) 6 (B) 12

C(s)

+

Control Systems

)

The angular frequency in (s )( sradians/s )( at ) (s s which the loop phase lag becomes is (A) 0.408 (C) 1.56 (B) 0.818 (D) 2.03

)

IN - 2009 6.

A plant with a transfer function

(

)

is

controlled by a PI controller with =1 and n un ty configuration. The lowest value of that ensures zero steady state error for a step change in the reference input is (A) 0 (C) 1/2 (B) 1/3 (D) 1

The steady state error for a unit step input when the gain = 1 is (C) 1 (A) (D) 2 (B)

IN - 2008 3. If a first order system and its time response to a unit step input are as shown below, the gain K is

Statement for the Linked data Q.No.7 & Q.No.8: A disturbance input (t)is injected into the unity feedback control loop shown in the figure. Take the reference input r(t) to be a unit step.

y(t)

r(t)

(C) 14 (D) 16

(s)

y

r(t)

+

d(t)

+ 1

+

1 S(S+1)

+

y(t)

0.8

7.

If the disturbance is measurable, its effect on the output can be minimized significantly (s) To using a feedforward controller eliminate the component of the output due to (t) s n t ( ) Should be (C) √ (A) √ (D) √ (B)

t

(A) 0.25 (B) 0.8

(C) 1 (D) 4

Statement for Linked Answer Questions 4 &5 A unity feedback system has open loop transfer function G(s) =

(

)

√

Let (s) ontroll r I (t) s n t the amplitude of the frequency component of y(t) due to d(t) is

8.

.The time

th

th

th


GATE QUESTION BANK

(A) √

(C) √

(B) √

(D) √

IN - 2011 11. The unit-step response of a negative unity feedback system with the open-loop transfer function G(s) = (A) (B)

IN - 2010 9. A unity feedback system has an open loop transfer function (s)

(

)

(s) (

(A) 0 (B) 0.5

(C) (D)

ECE/EE/IN - 2013 12. Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is

(A) u(t) (B) t u(t)

(s)

)

Y(s)

U(s)

A unit ramp input is applied to the system shown in the adjoining figure. The steady state error in its output is r(s)

is

.The value

of k that yields a damping ratio of 0.5 for the closed loop system is (A) 1 (C) 5 (B) 3 (D) 9 10.

Control Systems

(C)

u(t)

(D)

u(t)

(C) 1 (D) 2


(s [Ans. A] y(t) x(t) (s) (s) (

)

x(t)

utput

s s ( )u(t) When t = at steady state output = 1

√

√ s n(t)u(t)

y(t) 2.

s s

(t) (s)

s

√

s n .t

3.

/

[Ans. B] ,st p r spons s

[Ans. C] (t) (s) (s)

*

+

s s

4.

[Ans. B]

s

() (s) (s)

output s(s

)

)

s

(s

)s

( )

s th

th

s

.s th

/

s


GATE QUESTION BANK

t p r spons

s

.s

w

ns

s .s s

/

s s (t) ( )u(t) Which is also the required impulse response of the system.

/ (s

s

)

s

8.

When s =

[Ans. D] Comparing the given transfer functions with

⁄ (s)

5.

s

y(t)

s

,

-u(t) s

)(s

s

ξ

s

ln

ξ s Therefore P is undamped

[Ans. D] In dominant pole concept the factor that has to be eliminated should be in time constant form (s

Control Systems

s

s

ξ

)

s ov r / (s

. s

s

)

mp

s

s

ξ

s

s rt 6.

[Ans. C] ov rs oot

p n s on ξ s

M

r ξ os Where is the angle made by pole from negative real axis. To make M same, should be the same. 7.

s

√

[Ans. C] Transfer functions (s ) (s) (s )(s )

r

mp

9.

[Ans. A]

10.

[Ans. B] Steady state value = -2 Denominator ξ ξ underdamped

11.

[Ans. B] (s)

t

or

(s)

s

s un

s (s) (s) Output, (s) v n l m s (s) s (s ) (s )(s )

mp

ξ

Input (s)

or l m

lly

 1=

√

√



P=

√

(s ) (s )(s ) th

th

th


GATE QUESTION BANK

12.

(

[Ans. A] y y y x(t) t t Taking Laplace Transform of both sides (s) s (s) s (s) (s) (s) ( s ) (s) s s ol s r

ts

[Ans. C] (s) x(s)

15. (s

(s

)(s ) )(s )(s )

ξ Where damped natural frequency undamped natural frequency ξ damping ratio

y(s)

14.

√

16.

[Ans. C] ξ

(s s s pt on o

(s)

(s) y( ) 17.

(s)

u(t)

(s)

ξ So all poles of CLTP should lie on the left of s = line The characteristic equation is (s) (s) . We have to check whether are the roots this equation lies on me left of s = or not

o

[Ans. *] Range - 0.01 to 0.01 s (s) s x(t)

s ttl ng t m

pt on

[Ans. *] Range 38.13 to 38.19 We know √

s

(s )(s ) . / (s )(s )(s ) s For applying final value theorem system must be stable, mean all poles should lie in left half plane ) (s So (s ) r s

s)

(s )(s ) (s+1)(s+2)+0.15+5s=0 s s s s (s) pt on (s ) (s) (s) o (s ) (s )(s ) (s )(s ) (s ) s s s Hence Option C is correct

As poles are on the right hand side of splane so given system is unstable system. Only option (A) represents unstable system. 13.

Control Systems

s

(s)

s l m s y(s)

lm

s s

[Ans. *] Range 0.49 to 0.51 (s)

(s)

(s) s

( )

s (s) (s) )(s )(s s s (s)

s

)

(s) (s) (s) (s) (s) (s) (s) (s) (s) (s) * + (s) (s)

(s) (s)

(

s

)

(s) (s)

(s) th

th

(s) (s) th


GATE QUESTION BANK

Control Systems

(s) (s) r

(s)

s s

s s

2.

[Ans. C] m

s

l m s (s)

l ms

s s s

s s +

18.

i

[Ans. C] r ns r un t on or

or

s

ξ

s

s

r un t

∫ t t Taking Laplace transform

syst m

(s)

(

s

[Ans. A] Step input

s ∫ t

(s)

s

s

(s)

(s)

(s)

s

(s) (s)

s

t (s)

ξ s

s

(s) (s) (s)

(

s

)( s

ξ s

)

s

s

+

/

√

√

√

√ √

Overshoot /.

s

Comparing with s s

s .

(s)

*

s

(s) s Steady state value of Z t s (s) t

s

(s) s

.s

(s)

(s)

I(s)

Characteristic eq.

n (s) (s) (s) (s) (s) (s) (s) (s) (s) (s) (s) (s) (s) [ ] (s) (s) (s) (s) * + (s) (s) (s)

) I(s)

(s)

I(s) EE 1.

s

⁄ √

/

= 0.163 or 16.3% th

th

th


GATE QUESTION BANK

3.

[Ans. C]

5.

Control Systems

[Ans. C] M(s)

s s Comparing with standard form

√ √ r

M(s)

s

Setting time (t )

ξ

Square Wave

The system is critically damped 6.

[Ans. D] t y st t v lu o r spons Input is unit step So steady state error

T

For the square Wave ⁄ should be

ms For

r

(s)

(s)

n

w t

t

z ms

(s

t

t

Therefore, it would be best to have the as a square wave of 300Hz.

(s)

(s) (s)

ppl

)

(s)

s (s) (s) (s) s

t (

[Ans. A] r(t) un t mplus (t ) R(s) = 1[r(t)] = (s) (s) (s) s

)(s

s Steady state error using final value theorem t s (s)

z ms

(s) (s) (s)

(s)

rror

greater than t For z

4.

s

ξ

ms

For

s

ξ

)(

)

tt

s 7.

[Ans. C]

s s Steady state value of output, using final value theorem t s (s) s t s s

[

] ( s s rst or s .c Gain=2, T=1 sec For 98% t s

th

th

th

r syst m)


GATE QUESTION BANK

8.

[Ans. A]

R(s)

Control Systems

Given

C(s) +

G(s)

+

, 4.

[Ans. B] (s)

(

)

n t s (s) | (s) x(t)

,u(t)

x(s)

[

s (

s

 u(t s

)-

]

s

,

) | (s)

s

√

⁄

5.

[Ans. B]

6.

[Ans. A]

-

( ) ( )

IN 1.

t n ( ) t n ( ) r s -

d(t)= (t)

(s)

n

s or r(t) u(t) os t on l rror onst nt t (s)

( √ 8.

(s

s

] ) ( )

s(s (

)

(

)

(

)

+ ) )

(s)

, (s)

(s)

y(s)-

(

)

ontroll r

So s s s (s) * y(s) * + + ) s(s ) s(s y(s) s ( ) (s) s s t

)

t (s) [Ans. D] lm

(s)

(s) wor s s s s

)

s(s

) (

[Ans. B]

s (s)

)

⁄

y(s)

OR (s) (s)

( )

(t)

s nt *

(s)

3.

(

(t) [

[Ans. B] or

) (

( )

s s [Ans. D] Output due to

7.

( )

(

/ )

(

s m

( ) Phase lag It satisfy ,

(s)

/.

m

[Ans. D] (s)

2.

.

y(s) | | x(s)

s

th

th

√

th


GATE QUESTION BANK

9.

[Ans. D] T(s) =

(

)

= √

and

=

or 10.

Control Systems

√

or

K = 9.

[Ans. B] (s)

s(s ) For unit Ramp input, r(t)

t u(t)

lo ty rror onst nt (s) t

11.

[Ans. D] (s) (s) s The unit step response is given by (s) (t)

12.

)

s(s (

s

s

) u(t)

[Ans. B]

H(s) u(s)

⁄

Y(s)

u(s)= ⁄s [unit step i/p] ⁄s = ⁄ y(s) = H(s).U(s)= ⁄s s y(t) 0 ⁄s 1 t u(t)

th

th

th


GATE QUESTION BANK

Control Systems

Stability & Routh Hurwitz Criterion ECE - 2006 1. The positive values of and so that the system shown in the figure below oscillates at a frequency of r s respectively are R(s) (s

(s s

) s

C(s) )

4.

The number of open right half plane poles of G (s) =

is

(A) 0 (B) 1

(C) 2 (D) 3

ECE - 2012 5. The Feedback system shown below oscillates at 2 rad/s when (s)

(s s

(s

(A) 1, 0.75 (B) 2, 0.75

(C) 1, 1 (D) 2, 2

ECE - 2007 2. If the closed-loop transfer function of a control system is given as, (s) (A) (B) (C) (D)

(

)(

)

, then it is

an unstable system an uncontrollable system a minimum phase system a non-minimum phase system

, where

is a parameter.

Consider the standard negative unity feedback configuration as shown below +

)

ECE - 2014 6. The forward path transfer function of a unity negative feedback system is given by (s)

(s )(s ) The value of K which will place both the poles of the closed-loop system at the same location, is _______. 7.

Consider a transfer function (s)

(

)

(

)

with p a positive

real parameter. The maximum value of p until which remains stable is __________

G(s)

Which of the following statements is true? (A) The closed loop system in never st l or ny v lu o (B) or som pos t v v lu s o t closed loop system is stable, but not for all positive values (C) For all positive v lu s o t los loop system is stable (D) The closed loop system is stable for ll v lu s o ot pos t v n negative

(s)

(A) K=2 and a=0.75 (B) K=3 and a=0.75 (C) K=4 and a=0. 5 (D) K=2 and a=0.5

ECE - 2008 3. A certain system has transfer function G(s) =

) s

EE - 2007 1. The system shown in the figure is U1 +

s s

Σ

s

(A) (B) (C) (D)

th

Σ

U2 +

stable unstable conditionally stable stable for input u1, but unstable for input u2 th

th


GATE QUESTION BANK

2.

If the loop gain K of a negative feedback system having a loop transfer function ( (

) )

(C) unstable and of the minimum phase type (D) unstable and of the non-minimum phase type

to be adjusted to induce a sustained

oscillation then (A) The frequency of this oscillation must be ⁄ rad/s. √ (B) The frequency of this oscillation must be 4rad/s (C) The frequency of this oscillation

6.

The frequency response of the linear system ( ) is provided in the tabular form below. ( ) ( ) 1.3 1.2 1.0 0.8 0.5 0.3 The gain margin and phase margin of the system are (A) n (B) n (C) n (D) n

must be 4 or ⁄ rad/s √ (D) Such a K does not exist EE - 2008 3. Figure shows a feedback system where K > 0. The range of K for which the system is stable will be given by +

Σ

(A) 0 < K < 30 (B) 0 < K <39

s(s

)(s

)

(C) 0 < K < 390 (D) K > 390

EE - 2009 4. The first two rows of Routh's tabulation of a third order equation are as follows. s3 2 2 2 s 4 4. This means there are (A) two roots at s= ± j and one root in right half s-plane (B) two roots at s = ± j2 and one root in left half s-plane (C) two roots at s= ± j2 and one root in right half s-plane (D) two roots at s = ± j and one root in left half s-plane

Control Systems

EE - 2014 7. In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of (A) only one root at the origin (B) imaginary roots (C) only positive real roots (D) only negative real roots 8.

For the given system, it is desired that the system be stable. The minimum value of for this condition is __________ (s)

(s s

(

)s

(s)

) (

)s

(

)

EE - 2011 5. An open loop system represented by the transfer function G(s) = (

(

) )(

)

is

9.

A system wth the open loop transfer function

(A) stable and of the minimum phase type (B) stable and of the non-minimum phase type

(s)

th

s(s

th

)(s

s

th

)


GATE QUESTION BANK

is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is_________ 10.

A single-input single-output feedback system has forward transfer function G(s) and feedback transfer function H(s). It is given that |G(s)H(s)| < 1. Which of the following is true about the stability of the system? (A) The system is always stable (B) The system is stable if all zeros of G(s)H(s) are in left half of the s-plane (C) The system is stable if all poles of G(s)H(s) are in left half of the s-plane (D) It is not possible to say whether or not the system is stable from the information given

IN - 2006 1. The range of the controller gains (Kp, Ki) that makes the closed loop control system (shown in the following figure) stable is given as (s)

(s) s

(s

Ki  20 12 K (B) Ki < 0 and Kp  i  20 12

(A) Ki < 0 and Kp <

(C) Ki  0 and Kp  0 (D) Ki  0 and Kp 

Ki  20 12

)(s

)

Control Systems

IN - 2008 2. A closed loop control system is shown below. The range of the controller gain KC which will make the real parts of all the closed loop poles more negative than -1 is + s(s

(s) _

(A) KC> 4 (B) KC> 0

)

( )

(C) KC> 2 (D) KC< 2

IN - 2010 3. The open loop transfer function of a unity gain feedback system is given by: G(s) =

( (

) )(

)

. The range of positive

values of k for which the closed loop system will remain stable is: (A) 1
th

th

th


GATE QUESTION BANK

Control Systems


( ,

(s (

(s s From Routh array, s  k =2 & a = 0.75

) )

(

[Ans. D] System is stable, controllable and nonminimum phase system.

3.

[Ans. C] losed loop gain is ( ) (s) s s s Characteristic equation (s) s ( )s los loop system is stable only for r or or ll pos t v o the closed loop system is stable. [Ans. C] Perform Routh analysis to the polynomial in denominator of G(S). 1 3 5 s 2 6 3 s (∈) s (

s

.

)

(

s 1 a

[Ans. *] Range 2.24 to 2.26 Using root locus

X

X

Calculation of break point (s

)(s ) (s )(s ) s

s (s) , 7.

-

[Ans. *] Range 1.9 to 2.1 (s) to remain stable, the poles of For (s) should lie in LHP of s-plane 1 2–P s 3+P s 2–P s

/

[Ans. A] Characteristic equation is 1+G(S)H(S)= 0 (

= k + 1;

)

3 Number of sign changes =2

1+

6.

3

) (

s

)

(1+k) ( ) s s s = + (k + 1) = 0; a 4a = k + 1; From options, k = 2, a = 0.75

))

2.

5.

(

(s)

4.

)

o m x mum v lu o EE 1.

s

[Ans. D] u ( )

s s

)

+ s (k + 2) + (k + 1) = 0 k+2 k+2

y ( )

s th

th

th


GATE QUESTION BANK (

( ⁄ )

)

(

)

(

)

(s (s (

) )

5.

[Ans. B] For G(s), poles are at 2 and 3 Stable Also, zeroes are to the right of S-plane non-minimum phase type

6.

[Ans. A] At gain across over frequency ( magnitude of ( ) is 1. | ( )|

)

Pole is in LHS of s-plane, Hence stable. u ( )

y ( ) s

s s

( ⁄ ) (

)

Phase of (

(

)

(

)

(

)s

M

(

)p

s o

log log

)

(

)

1

s

7.

[Ans. B] It is a special case under R-H stability criteria. Imaginary roots are calculated using auxiliary polynomial

8.

[Ans. *] Range 0.61 to 0.63 Characteristic equation ( )s ( )s ( s

64+3k

s 64+3k (or)

for marginal stability

For K = 16,

r

⁄s

s

[Ans. C] (s)

( ( )( ) Characteristics equation s s s s s s

s s s

)

( )s rt r 1

[Ans. D] s s  (s

s )(s

)

)

s

s

1 (

)

1 )

s (

( (

) )(

)

√ √

& k < 390 (for stability) 4.

)

) ( ) t

m gn tu o | ( )|

Gain margin=

),

) (

(

[Ans. B] T(s) =

(

At phase cross frequency (

(s )(s ) Hence unstable as it has pole at right hand side of S-plane

3.

)

Phase of margin =

s

2.

Control Systems

S= th

th

th


GATE QUESTION BANK

Control Systems

s

√

[Ans. 5] Characteristic equation )(s ) s(s s (s ) s)(s s s s s s s s s s s

IN 1.

s

(

s s s s

1 4 5

s

1

)

6 4 K

n  s 2.

[Ans. C] ) s(s s s

k 0

√

√

√

[Ans. A] Characteristic equation of system (s) (s) G(s) H(s) = For system to be stable all the zeroes of G(s) H(s) should lie in left of ( ) plane So for (s) (s) The system will be always stable.

or √

Put S = Z – 1 and apply RH criterion for the polynomial in z

Value of k to be marginally stable K=5 10.

12

√

M n mum v lu 9.

s √

3.

[Ans. D] 1 + G(s) H(s) = 0 (s + 1) (s + 2) + k(s+3) = 0 System is stable, for all positive K.(from Routh Hurwitz criterion) (or)

4.

[Ans. D] 3 4 S 0(∈ 1 4

[Ans. D] Characteristic equation is s(s+2)(s+10)+( ) s s s

3 4 ) s

(

)(

)

S=

th

th

th


GATE QUESTION BANK

Control Systems

Root Locus Technique ECE - 2007 1. A unity feedback control system has an open-loop transfer function (s)

(

)

. The gain K for which

s= 1+j1 will lie on the root locus of this system is (A) 4 (C) 6.5 (B) 5.5 (D) 10 ECE - 2009 2. The feedback configuration and the polezero locations of G(s) =

are shown

(A)

(s) (s)

(B)

(s) (s)

(C)

(s) (s)

(

(

)(

( (

)

)( (

(s) (s)

(D)

) )( ) ( )

(

)(

)

) )

)

ECE - 2014 4. Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown. Which one of the following conclusions is correct? (s)

below. The root locus for negative values of k, i.e. for s breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

Im (

)

+ +

k

G(s)

(

Im(s)

(A) G (s) is an all-pass filter (B) G (s) is a strictly proper transfer function (C) G (s) is a stable and minimum-phase transfer function (D) The closed-loop system is unstable for sufficiently large and positive k

Re(s) P

(A) ± √ and 00 (B) ± √ and 450

(C) ± √ and 00 (D) ± √ and 450

)

5.

In the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of the following transfer functions has this root locus?

ECE - 2011 3. The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

th

th

th


GATE QUESTION BANK

(A) (B) (C) (D) 6.

(

)(

(

)(

)(

( (

)( )(

)( )

(

)(

)

j

(C)

)

)(

Control Systems

) )

2

The characteristic equation of a unity negative feedback system is (s) . The open loop transfer function G(s) has one pole at 0 and two poles at 1. The root locus of the system for varying K is shown in the figure.

+1

1

(D)

+2

j

+1

1

2

+2

ξ

(

)

The constant damping ratio line, for ξ nt rs ts t root lo us t po nt The distance from the origin to point A is given as 0.5. The value of K at point A is__________ EE - 2006 1. A Closed – loop system has the characteristic function (s )(s ) + K (s ) = 0. Its root locus plot against K is j𝛚 (A)

2

(B)

1

EE - 2010 2. The characteristic equation of a closedloop system is )(s ) (s ) s(s . Which of the following statements is true? (A) Its roots are always real (B) It cannot have a breakaway point in ,s- 0 the range (C) Two of its roots tend to infinity along the asymptotes Re, (D) It may have complex roots in the right half plane. EE - 2011 3. The open loop transfer function G(s) of a unity feedback control system is given as, G(s) =

j

2

+1

/

(

)

From the root locus, it can be inferred that when k tends to positive infinity. (A) three roots with nearly equal real parts exist on the left half of the splane (B) one real root is found on the right half of the s-plane (C) the root loci cross the j axis for a fin t v lu o ≠ (D) three real roots are found on the right half of the s-plane

+2

+1

.

+2

th

th

th


GATE QUESTION BANK

EE - 2014 4. The root locus of a unity feedback system is shown in the figure

Control Systems

controller of gain K in the forward path, in a unity feedback configuration. The transfer function is (A) ( )( )( (C) ( )( )( )

2.

)

X

(B)

X

3. The closed system is (s) ( ) (s) (s) ( ) (s) (s) ( ) (s) (s) ( ) (s)

)(s

)

(s

)(s

)

(s

)(s

)

(s

)(s

(

)(

)(

)

( (

) )(

)

.

The root locus plot of the system has (A) Two breakaway points located at s = 0.59 and s = 3.41 (B) One breakaway point located at s = 0.59 (C) One breakaway point located at s = 3.41 (D) One breakaway point located at s = 1.41

imaginary at ω=4 2 rad/s with gain K=384. It is observed that the point s= 1.5+j1.5 lies in the root locus. The gain K at +j1.5 is computed as

IN - 2009 5. A unity feedback system has the transfer

1.5+j1.5

(A) 11.3 (B) 21.2

(D)

)

feedback system is G(s)=

)

X 4

)(

IN - 2008 4. The open loop transfer function of a unity

IN - 2006 1. The root locus of a plant is given in the following figure. The root locus crosses

X 8

)(

If the root locus plot of the closed loop system passes through the points ± j√ , the maximum value of K for stability of the unity feedback closed loop system is (C) 10 (A) √ (B) 6 (D) 6√

loop transfer function of the

(s

(

X

function

0

) )

. The value of b for which

the loci of all the three roots of the closed loop characteristic equation meet at a single point is (A) 10/9 (C) 30/9 (B) 20/9 (D) 40/9

(C) 41.25 (D) 61.2

IN - 2007 Statement for Linked Answer Questions 2 and 3 A transfer function with unity DC gain has three poles at and and no finite zeros. A plant with this transfer function is connected with a proportional

( (

IN - 2011 6. Consider the second-order system with the characteristic equation ) (s ) s(s . Based on the properties of the root loci, it can be shown that the complex portion of the root loci of the given system for is described by a circle, and the two breakaway points on the real axis are th

th

th


GATE QUESTION BANK

(A)

√

(B)

√

(C)

√ √

(D)

IN - 2012 7. The open loop transfer function of a unity gain negative feedback control system is given by G(s) =

(

)(

)

ngl

Control Systems

IN - 2014 8. A loop transfer function is given by : (s ) (s) (s) ( ) The point of intersection of the asymptotes of G(s)H(s) on the real axis in the s-plane is at ___________.

t

which the root locus approaches the zeros of the system, satisfies (A)

t n

. /

t n

. /

(B)

=

(C)

=

t n

. /

(D)

=

t n

. /


[Ans. D] ro u t o p sor r wn rom op n loop pol s to t t po nt pro u t o p sor r wn rom op n loop z ro to t t po nt √ √

√

2.

√

[Ans. B] ( (

5. ) )





[Ans. B]

√

Angle of departure is Where Σ z Σ z

3.

4.

[Ans. B] From plot we can observe that one pole terminates at one zero at position and three poles terminates to . It means there are four poles and 1 zero. Pole at goes on both sides. It means there are two poles at

There are 4 poles/zeros at s = 1, There has to be one pole – zero pair at both cannot be poles or zeros at the same time. So option A and D are crossed out There are 2 asymptotes. So number of poles is two more than number of zeros.

[Ans. D] For larger values of K, it will encircle the critical point ( ) which makes closed-loop system unstable. th

th

th


GATE QUESTION BANK

(s

Since there has to be one zero at either s or s , the remaining break frequencies are all poles ons r opt on ntro

po nt w

r t

)

(s )(s ) Zero of OLTF s = 1: z = 1 Poles of OLTF s =

symptot s m t

The picture forms to this For option C, centroid is at 0 which is wrong Hence Option B is correct. 6.

Control Systems

s

l n

The root locus starts from open – loop poles and terminates either on open loop zero or infinity. Root locus exist on a section of real axis it is the sum of the open – loop poles and zeros to the right of the section is odd. Number of branches terminating on infinity.

[Ans. *] Range 0.32 to 0.41

ξ

Angles of asymptotes ( ) ( ( )

(

or

)

Intersection of asymptotes on real axis (centroid) Σpol s Σz ros

So the co-ordinates of point A is √ On the root locus, we know

(

( )

√

[Ans. B] Characteristic function (s )(s ) (s ) (s ) (s) (s) (s )(s ) Open loop transfer function (s) (s)

)

( )

1 Option (B) is correct on the basis of above analysis.

( ) EE 1.

)

2.

[Ans. C] Characteristic equation )(s ) (s ) s(s (s ) s(s )(s ) Comparing with 1 + G(s) H(s) = 0 G(s)H(s) = open-loop transfer function th

th

th


GATE QUESTION BANK

(

(s

)

)

)(s ) s(s Number of zeros = Z = 1, zero at Number of poles = P = 3 Poles at 0, & Number of branches terminating at infinity = Angle of asymptotes ( ) (

)

(

)

Control Systems

2

0

⁄

So all these roots have nearly equal real parts on left half plane when 4.

[Ans. C]

n Σ pol s

ntr o

(

Σ z ros

) X

X

Im g n ry

X

X

Given root locus is complementary root locus for which either k < 0 or loop gain is +ve means +ve feedback, means G(s) is always +ve So for k > 0

l

G(s) H(s) =

Breakaway point lies in the range ,sand two branches terminates at infinity along the asymptotes Re(s) 3.

[Ans. A] (s)

(s s (s

(s) (s) IN 1.

)(

)

(s)

s) )

(s

)(s

s(s (s)

)

)(s (s)

(s)

Angle of asymptotes

(s) (s) (s)

[Ans. C]

ntro

q

(

(s) (s)

)

s s

s

ts

…( – Z – 1)

(

s

√

)

√

s

√

for poles at zero , -

s

√ √ √ √

so Rool locus is

th

th

√

th


GATE QUESTION BANK

Note that the data given at the intersecting point with imaginary axis is not necessary 2.

3.

(s

s ) (s ) We need to find the breakaway point. o

[Ans. B] With unity DC gain, poles at s n and no finite zeros l nt

(s

)(s

)(s

s )( s s) (s s ) s s s s s ( )s s s ow s is not the breakaway point ( )s o s For all the three root loci to meet at a single point, we need that this equation has equal roots. ) o( (s s

)

( ) (

( )

 s s s s

s

)(

)(

(

s

)

) (

(

)

)

( ) row has zero elements if 6k = 60 or k = 10 then 2 roots lie on the imaginary axis given by s s √ . If k > 10, the root at √ s t to o s pl n 4.

(

(

)

(



)

(



so

(

) (

(

)

) (

)

)

( ) or

o

st

r qu r

.(

)(

6.

[Ans. C] K=

(

)

(

) ) ( (

7.

√

s s(s )(s (s ( s(s

(s)

)

[Ans. B] (s ) s (s ) For unity feedback, equation is 1+ G(s) = 0 s s s

√

[Ans. D]

(s) 5.

) /

)

)

S=

(

v lu

=0

s s S= √ RL exists at 3.41 (

)

But b = 20 is not the required value of b because it will cancel out an open-loop pole

) (

) )(

(

[Ans. C] (s)

s

or

[Ans. C] T(s)=

Control Systems

) ))(s ( )(s )

))

(s)

characteristic

th

th

th


GATE QUESTION BANK

Control Systems

Angle at which RLD approaches to zeros is the angle of arrival for zeros. , (

)

0

. /1

t nt [ * [

( )

t n t nt t n

8.

( )] ( )]+ ( )

[Ans. ] Int r s t on o symptot s sum o pol s sum o z ros num r o pol s num r o z ro ( ) ( )

th

th

th


GATE QUESTION BANK

Control Systems

Frequency Response Analysis Using Nyquist plot ECE - 2006 1. The open-loop transfer function of a unity-gain feedback control system is given by, G(s) =

(

)(

. The gain

)

margin of the system in dB is given by (A) 0 (C) 20 (B) 1 (D) 2.

Consider

two

(s)

and

transfer (s)

G(s)

4.

Which of the following statement is true? (A) G(s) is an all-pass filter (B) G(s) has a zero in the right-half plane (C) G(s) is the impedance of a passive network (D) G(s) is marginally stable

5.

The gain and phase margins of G(s) for closed loop stability are (A) 6 dB and (C) 6 dB and (B) 3 dB and (D) 3 dB and

functions, . The

3-dB bandwidths of their frequency responses are, respectively

3.

(A) √

√

(B) √

√

(C) √

√

(D) √

√

The Nyquist plot of ( ) ( ) for a closed loop control system, passes through ( ) point in the GH plane. The gain margin of the system in dB is equal to (A) Infinite (B) greater than zero (C) less than zero (D) zero

ECE - 2011 Common Data Question 6 and 7: The input – output transfer function of a ( )

plant

(

)

. The plant is placed

in a unity negative feedback configuration as shown in the figure below. r

u

(s)

y s(s

)

l nt

ECE - 2009 Common Data for Question 4 and 5 The Nyquist plot of a stable transfer function G(s) is shown in the figure. We are interested in the stability of the closed loop system in the feedback configuration shown

6.

The signal flow graph that DOES NOT model the plant transfer function ( ) is (A) 1

u

1/s

1/s

1/s

100 y

Im

(B)

u

1/s

1/s

1/s

100

y

Re

th

th

th


GATE QUESTION BANK

Control Systems

(D)

(C)

1/s

u

1/s

1/s

100

y

⁄

(D) 1/s

1/s

u

7.

1/s

100

y

The gain margin of the system under closed loop unity negative feedback is (A) 0 dB (C) 26 dB (B) 20 dB (D) 46 dB

EE - 2006 1. Consider the following Nyquist plots of loop transfer functions over = 0 to = . Which of these plots represents a stable closed loop system? (1)

Im R =

8.

For the transfer function ( ) the corresponding Nyquist plot for the positive frequency has the form (A)

Re 1

Im

(2)

5

=

Re

(B) (3)

Im

j5

= Re

(C)

(4)

Im

= ⁄

1

(A) (B) (C) (D)

th

Re

(1) only all, except (1) all, except (3) (1) and (2) only

th

th


GATE QUESTION BANK

EE - 2007 2. If x = Re G(j ), and y = Im G(j ) then for , the Nyquist plot for G(s) = 1 / s(s+1) (s+2) becomes asymptotic to the line (A) x = 0 (C) x = y (B) x =

Control Systems

(A)

Im 3/4 Re

(D) d = y / √

EE - 2009 3. The polar plot of an open loop stable system is shown below. The closed loop system is

𝛚=0 𝛚=0

(B)

Im

Imaginary

Real

Re 3/4

(A) Always stable (B) Marginally stable (C) Unstable with one pole on the RHS splane (D) Unstable with two poles on the RHS s-plane 4.

(C)

The open loop transfer function of a unity feedback system is given by ( )⁄ G(s) = s The gain margin of this system is (A) 11.95dB (C) 21.33dB (B) 17.67dB (D) 23.9dB

𝛚=0

Im

Re

1/6

Im

(D)

EE - 2010 5. The frequency response of (s) )(s )- plotted in the ,s(s complex ( ) plane (for ) is

Re

𝛚=0 1/6

th

th

th


GATE QUESTION BANK

EE - 2011 6. A two-loop position control system is shown below. Motor s(s

IN - 2009 4. A unity feedback control loop with an open transfer function of the form

(s)

(s)

having a transfer function

s Tacho-generator

by G(s) = ( 5.

G(s) = 2.

3.

(

)(

6.

The gain margin of the system is (A) 0.125 (C) 0.5 (B) 0.25 (D) 1

(C) 125√ (D) 125√

v lu o or t mp ng r t o to be 0.5, corresponding to the dominant closed-loop complex conjugate pole pair is (A) 250 (C) 75 (B) 125 (D) 50

IN - 2012 7. The open loop transfer function of a unity negative feedback control system is given by G(s) =

)

The phase crossover frequency of the system in radians per second is (A) 0.125 (C) 0.5 (B) 0.25 (D) 1

is inserted

)

(A) 250√ (B) 250√

⁄

Statement for Linked Answer Questions 2 &3 Consider a unity feedback system with open loop transfer function

√

The value of K for the phase margin of the system to be 45° is

(s)

(C) (D)

√

IN - 2011 Common Data for Questions 5 and 6 The open-loop transfer function of a unity negative feedback control system is given

IN - 2008 1. For the closed loop system shown below to be stable, the value of time delay TD (in seconds) should be less than

⁄ ⁄

)

into the loop, the phase margin will become (A) (C) (B) (D)

The gain k of the Tacho-generator influences mainly the (A) peak overshoot (B) natural frequency of oscillation (C) phase shift of the closed loop transfer function at very low ) frequencies ( (D) phase shift of the closed loop transfer function at very high ) frequencies (

(A) (B)

(

has a gain crossover frequency of 1 rad/s and a phase margin of . If an element

)

(s)

Control Systems

(

)(

)

. The gain margin of the

system is (A) 10.8 dB (B) 22.3dB

(C) 34.1dB (D) 45.6dB

IN - 2014 8. The loop transfer function of a feedback control system is given by (s) (s)

)( s ) s(s Its phase crossover frequency (in rad/s), approximated to two decimal places, is __. th

th

th


GATE QUESTION BANK

Control Systems


6.

[Ans. D] 𝐹or option (D).

[Ans. D]

( )

(s)

(s )(s or g n m rg n t n ( ) t n

(

( )

)

t n

⁄

. /

7.

[Ans. C]

)

(s) (s)

) s(s ( ) t n phase cross-over

=0 o (s)

For

[ ]

Gain margin =

log

t n (

0 1 r

Using Root locus of G(s)

) (

(

root locus will cut the when k

) (

3.

M

= 0 axis only 0 1

8.

[Ans. D] At ( GM = 20 log

4.

[Ans. B]

5.

[Ans. C] ( At log

|

(

) (

)

)

( ) (

)

)

log

[Ans. A] Assuming no.of open loop poles in the RHS of s – plane = P = 0 Complete nyquist plots

|

Im

) (

)

GM = 20

| (

At

|

EE 1.

)

)

[Ans. A] ( ) =5 So ( ) is a straight line parallel to axis.

)

) (

(

)

(

M n

[Ans. C] BW depends only on denominator and is equal to √(

)

s

)

(

So gain margin =

frequency,

ut s (

2.

(s)

s not tr ns r un t on o

) (

)

PM = No.of encirclements = N = 0 th

th

th


GATE QUESTION BANK

N=P–Z=0

Control Systems

Hence the system is unstable So option (A) is correct

Hence system is stable.

2.

[Ans. B]

Imj

(

) (

)

( (

)(

symptot to x = 3.

Two anti – clockwise encirclement N=2 P–Z=N=2

(

)

) )(

)

s

[Ans. D] 2 clockwise encirclement of 1 + j0 = 0; N = 2 =2 Z = number of closed loop ploes in RHS Hence system is unstable

4.

[Ans. D] Open loop transfer function, G(s) = Put s =

Hence system is unstable. Im

(

)

At phase crossover frequency ( phase of OLTF is ( )

Two clockwise encirclement of – 1 Hence N = 2

|

| s lw ys (

(

Im

or ny v lu o

) n t

Hence the system is unstable

)

(p

s

ross r qu n y)

)

Gain margin = 20log

| (

)|

) )

((

log 5.

[Ans. A] (

)

As

( ((

)

(

G(

) )

(

))

)

s ( ) -0 + Option (A) satisfies above

Two clockwise encirclement of – 1 Hence N = 2 Z=2 th

th

th


GATE QUESTION BANK

6.

[Ans. A] T(s) =

2. (

)

r

s

(

(

)

(

ξ

)

)

M is dependent on K (A) is correct is independent of K (B) is not correct T(J ) = (

[Ans. C] ( ) t n ( ) 

)

(

)

(

( )

Control Systems





3.

[Ans. A] ( )

4.

[Ans. A] M o (s)

r

t n

.

(

)

s

/

( )≈ ( )≈

( ) n ( ) r p rt lly orr t

( [Ans. C]

(s) = s

(s)

)

)

t low r qu n s t g r qu n s

IN 1.

t n ( )

t n (

( (

PM of

y(s)

5.

t n √ ) √ )

(

√

)

(s) =

(

)

[Ans. B] PM = 45°= 180° t n

(s)

. /) 

t n

. /

rad/sec

( )

y(s)

. /

|(

)

|

√ G(s) = For system to be stable G(s) should lie in the left of ( 1, 0) For finding the critical/ marginal value apply phase condition w

6.

[Ans. A] r t r st

(s ) s s s ( The dominant poles are given by ) s s (

w

or (s) (s) w

qu t on

s

s

(

)

)

s |

s

|

w

So w

For stable system

or

7.

[Ans. C] = 15 r/s √ th

th

√ th


GATE QUESTION BANK

G.M in dB = 8.

log 0

Control Systems

1

[Ans. *]Range 0.30 to 0.34 (s) (s)

s(s

)( s

)

or (s) (s)

t n

t n ( )

(

t n (

)

)

As t n So

=1 r s r s

th

th

th


GATE QUESTION BANK

Control Systems

Frequency Response Analysis Using Bode Plot ECE - 2006 Statement for Linked Answer Questions 1 &2 Consider a unity-gain feedback control system whose open-loop transfer function is

ECE - 2010 4. For the asymptotic Bode magnitude plot shown below, the system transfer function can be M gn tu

( ) 1.

The value of 'a' so that the system has a phase-margin equal to is approximately equal to (A) 2.40 (B) 1.40

2.

(C) 0.84 (D) 0.74

With the value of 'a' set for a phasemargin of the value of unit-impulse response of the open-loop system at second is equal to (A) 3.40 (C) 1.84 (B) 2.40 (D) 1.74

(A)

(C)

(B)

(D)

ECE - 2014 5. The phase ( )

(

margin )(

)(

in )

degrees

calculated

of

using

the asymptotic Bode plot is ________. ECE - 2007 3. The asymptotic bode plot of a transfer function is as shown in the figure. The transfer function G(s) corresponds to this Bode plot is |G(

6.

In a Bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4th order all-pole system (A) (C) (B) (D)

7.

The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

)|(dB)

60 20 dB/decade 40 40 dB/decade 20

0

0.1

1

10

2 0

( (

100

) )

60 dB/decade

(A) (B)

(

)( (

)(

(C)

) )

(D)

(r (

)(

(

)(

)

⁄s) n log s l

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is_________.

)

th

th

th


GATE QUESTION BANK

EE - 2006 1. The Bode magnitude plot of ( )(

H (j ) = ( (

(A)

) )

Control Systems

EE - 2008 2. The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure.

is

)

( 40

)

40 dB/ decade

(dB) 20

20

20 dB/ decade (r s) (log scale)

0.1 0

+1

+3

+2

Log ( )

20

0 dB/ decade

0

(B)

(

This transfer function has (A) Three poles and one zero (B) Two poles and one zero (C) Two poles and two zeros (D) One pole and two zeros

) 40 20

0

+1

+2

+3

Log ( )

EE - 2009 3. The asymptotic approximation of the log-magnitude vs frequency plot of a system containing only real poles and zeros is shown. Its transfer function is 40 dB / dec

(C)

(

)

60 dB / dec

80 dB 40

20

0

+1

+2

+3

r

Log ( )

0.1

(A)

(D)

( )(

(

(B) (

2 5

(

) ( )(

(C)

) )

( )( (

(

(D)

)

s

25

(

) ) ) )(

)

) 40

EE - 2014 4. The Bode magnitude plot of the transfer

20

function (s) 0 +1

+2

+3

( .

)( /(

) ).

/

is shown

below: Note that -6 dB/octave = -20 dB/decade. The value of is_______

Log ( )

th

th

th


GATE QUESTION BANK

The gain and phase margin of the system are (A) 0 dB, 180° (C) 0 dB, 10° (B) 3.88 dB, 170° (D) 3.88 dB, 10°

t v t v

Control Systems

t v t v

t v

t v (r

⁄s )

IN - 2008 2. The Bode asymptotic plot of a transfer function is given below. In the frequency range shown, the transfer function has dB

5.

6.

For the transfer function (s ) ( ) s(s )(s s ) The values of the constant gain term and the highest corner frequency of the Bode plot respectively are (A) 3.2, 5.0 (C) 3.2, 4.0 (B) 16.0, 4.0 (D) 16.0, 5.0

+ 20dB / decade 20dB / decade 0dB / decade

log

(A) (B) (C) (D)

The magnitude Bode plot of a network is shown in the figure (

) lop

log

The maximum phase angle and the corresponding gain respectively, are (A) 30° and 1.73dB (B) 30° and 4.77dB (C) + 30° and 4.77dB (D) + 30° and 1.73dB IN - 2006 1. A unity feedback system has the following open loop frequency response: (r s ) ( ) ( ) 2 7.5 3 4.8 4 3.15 5 2.25 6 1.7 8 1 10 0.64

3 poles and 1 zero 1 pole and 2 zeroes 2 poles and 1 zero 2 poles and 2 zeroes

IN - 2010 3. The asymptotic Bode magnitude plot of a lead network with its pole and zero on the left half of the s-plane is shown in the adjoining figure. The frequency at which the phase angle of the network is maximum (in rad/s) is

r s (log scale)

(A) (B)

(C)

√

(D)

√

IN - 2013 4. The Bode plot of a transfer function G(s) is shown in the figure below.

th

th

th


GATE QUESTION BANK

Control Systems

Gain (dB)

The gain (20 log|G(s)|) is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all . Then G(s) is 40 32

(A)

(C)

(B)

(D)

20

5.

The discrete – time transfer function is

0 1

10

-8

(r

s)

(A) (B) (C) (D)

100

Non – minimum phase and unstable. Minimum phase and unstable. Minimum phase and stable. Non – minimum phase and stable.


4.

[Ans. A]

[Ans. C] | ( )| M w r

G(s) = ( (

.

/

(

)

.

/

(

)

)

)

t n (

5.

)

[Ans. *] Range 42 to 48 (

)

M t n ( √

2.



) 

√



= 0.84

[Ans. C] (s)

(

)

 g(t)

u(t)

t u(t)

(

3.

[Ans. D] (s) o

s(s

)(

plot s ( r t’s typ log

) ) orm syst m so

)

( )( )( ) From the plot frequency at which gain is 0 dB is r s ( ) M

At t = 1, g(t)

6.

[Ans. A] In a transfer function if all are poles if we plot the BODE diagram, then an each and every corner frequency we have to introduce a line of slope and hence on the 4th frequency the slope of line will become – 80 dB/sec and will continue upto infinity

|

| K = 100

th

th

th


GATE QUESTION BANK

7.

[Ans. *] Range 0.49 to 0.51 From the Bode plot given, we can see that there are 2 break frequencies at = 2 and = 10 rad/sec. Also it is a type 1 system. So, the transfer function is ( ) ( ) ( ) ( ) At = 1 rad/sec, (given)

2.

[Ans. C] Initial slope is – 40 dB/decade, it means there are double pole at origin. Slope changes from dB/decade to dB/decade. It means there is a zero. Slope changes from dB/decade to 0 dB/decade at some other frequency. It means there is one more zero. Therefore transfer function has two poles and zeros.

20 log | |

3.

[Ans. B]

log ( ( In s domain, ( (s) ( (

) )

)

l m s (s) (

log

s) s)

(s s(s (s lm s

) ) )

/

) )

[Ans. *] Range 0.7 to 0.8 For initial dotted slope

[Ans. A] (

/

/.

(s )(s

s (s

( )(

log )

s .

)

)

( .

4.

(

( s ) s )( s )

s (

So,

EE 1.

Control Systems

(

)

)

log

)

( )

log (

( )( ) Corner frequencies are 1, 10 & 100 rad/sec ( ) ( )

log

)

log .

(s) .

log

/.

/.

/

/.

/

log (

)

5.

[Ans. A] (s)

(s )(s

s(s (

log

)(

s( log log log

log log log

(

) s

s )

s) .

/

)

n m x mum r qu n y th

)

th

th

r s


GATE QUESTION BANK

6.

Control Systems

[Ans. C]

.

(

/

(s)

)

.

/

has 2 poles and 1 zero s s

3.

t n

[Ans. B] This condition is only for phase lead N/W

t n

=√ (

4.

)

(

)

(

√

[Ans. B] 32 dB

)

1

10

√ t n √

. /

√

√ √

√

g n

to Is 1 dec are change & change is (G) is 40dB lop s r r pol s s or g n

√

. / √

g n IN 1.

t n

√

log √

So, G(s)= = 32 dB (given)

[Ans. D]  M r s  M

log

(

=

) 5.

[Ans. D] z z For minimum phase system, all poles and zeros must lie inside the unit circle. For stable system, all poles must be inside the unit circle. For the system, zero is at 2 pole is at 0.5. This system is stable but non – minimum phase.

[Ans. C] Compare with Bode magnitude plot of standard transfer function. . (

(z)

/

) .

. /|

log

= 3.88 dB 2.

log

/

th

th

th


GATE QUESTION BANK

Control Systems

Compensators & Controllers ECE - 2006 1. The transfer function of a phase-lead

Z -

(s)

compensator is given by

+

where . The maximum phase-shift provided by such a compensator is (A) ⁄ (C) ⁄ (B) ⁄ (D) ⁄

Group I

ECE - 2007 2. A control system with PD controller is shown in the figure. If the velocity error constant Kv =1000 and the damping ratio ξ t n t v lu s o p and KD are r

(A) (B) (C) (D) 3.

+

Σ

s

s(s

)

Kp=100, KD=0.09 Kp=100, KD=0.9 Kp=10, KD=0.09 Kp=10, KD=0.9

The open-loop transfer function of a plant is given as 0 (s)

1. If the plant is

operated in a unity feedback configuration, Then the lead compensator that can stabilize this control system is (A) (B)

(

)

(

)

(C) (D)

(

)

(

)

Group II 1. PID controller 2. Lead compensator 3. Lag compensator (A) Q – 1, R – 2 (C) Q – 2, R – 3 (B) Q – 1, R – 3 (D) Q – 3, R – 2

y

ECE - 2009 5. The magnitude plot of a rational transfer function G(s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot? (

) 20dB

log

ECE - 2008 4. Group 1 gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the condition R2C2 >R1C1. The transfer function represents a kind of controller. Match the impedances in Group I with the types of controllers in Group II.

(A) (B) (C) (D)

Lead compensator Lag compensator PID compensator Lead – lag Compensator

ECE - 2010 6. A unity negative feedback closed loop system has a plant with the transfer function G(s) =

and a controller

(s) in the feed forward path. For a unit step input, the transfer function of the

th

th

th


GATE QUESTION BANK

controller that gives minimum steady state error is (A)

(s)

(B)

(s)

(C)

(s)

(D)

(s)

(

)(

)

(

)(

)

s

ECE - 2012 Statement Linked answer Questions 7 and 8 The transfer function of a compensator is given as (s) . 7.

8.

Gc(s)is a lead compensator if (A) (B) (C) (D)

Control Systems

(D) A lag-lead compensator that provides an attenuation of 20 dB and phase lead of 450 at the frequency of 3 rad/s EE - 2008 2. The transfer function of two compensators are given below (s ) s (s ) (s ) Which one of the following statements is correct? (A) C1 is a lead compensator and C2 is a lag compensator (B) C1 is a lag compensator and C2 is a lead compensator (C) Both C1 and C2 are lead compensators (D) Both C1 and C2 are lag compensators IN - 2006 1. The transfer function of a position servo

The phase of the above lead compensator is maximum at (C) √ r s (A) √ r s (D) s (B) √ r s √ r

system is given as G (s) =

1 . A first s(s  1)

order compensator is designed in a unity feedback configuration so that the poles of the compensated system are placed at – 1  j1 and 4. The transfer function of the compensated system is

EE - 2007 1. The system 900/s(s+1)(s+9) is to be compensated such that its gain-crossover frequency becomes same as its uncompensated phase-crossover frequency and provides a 450 phase margin. To achieve this, one may use (A) A lag compensator that provides an attenuation of 20 dB and a phase lag of 450 at the frequency of √ rad/s (B) A lead compensator that provides an amplification of 20dB and a phase lead of 450 at the frequency of 3 rad/s (C) A lag-lead compensator that provides an amplification of 20 dB and a phase lag of 450 at the frequency of √ rad/s.

(A)

(

(C)

)

(D)

(B)

(

)

(

)

Common Data Questions 2, 3, 4 The following figure describes the block diagram of a closed loop process control system. The unit of time is given in minute ontroll r (s)

r(s)

th

th

(

)

th

stur

l nt

n

(s)

m(s) s


(s)

GATE QUESTION BANK

2.

The digital implementation of the controller with a sampling time of 0.1 minute using velocity algorithm is ( ) m( )

* ( )

∑ ( )+

( ) m( )

* ( )

∑ (

( ) m( )

m(

( ) m( )

m(

4.

motor is given as

( ) ( )

. When

connected in feedback as shown below, the approximate value of that will reduce the time constant of closed loop system by one hundred times as compared to that of the open – loop system is

)+

(

)-

(

)-

R(s)

(s)

(s)

) ( )

, 3.

ECE/EE/IN - 2013 6. The open – loop transfer function of a dc

) , ( )

Control Systems

Suppose a disturbance signal d(t)= sin 0.2t unit is applied. Then at steady state, the amplitude of the output e(t) due to the effect of disturbance alone is (A) 0.129 unit (C) 0.529 unit (B) 0.40 unit (D) 2.102 unit

(A) 1 (B) 5

(C) 10 (D) 100

IN - 2014 7. Consider the control system shown in figure with feed forward action for rejection of a measurable disturbance d(t). The value of K, for which the disturbance response at the output 𝑦( ) is zero mean, is:

The control action recommended for reducing the effect of disturbance at the output(provided that the disturbance signal is measurable) is (A) cascade control (B) P-D control (C) ratio control (D) feedback-feed forward control

(t)

y(t)

r(t)

s

(A) 1 (B) 1

IN - 2007 5. A Cascade control system with proportional controllers is shown below.

(C) 2 (D) 2

(s) (

)(

)

( s

)

(s)

Theoretically, the largest values of the gains and that can be set without causing instability of the closed loop system are: (A) 10 and 100 (C) 10 and 10 (B) 100 and 10 (D) n

th

th

th


GATE QUESTION BANK

Control Systems


4.

[Ans. B]

[Ans. D] M xp s s t (s) t n t n From maximum phase shift

s s s

t ( (

,

( (

) ) -

(

(t ) ) s (s) (s )(s ) (s) s The above equation is a PID controller

)

)

or

√ t n (√ ) 2.

t n

√

s ( ) ( ) ( ) ( ) The above equation is a lag compensator. ( )

[Ans. B] (s)

( (s m m

)

s (s) (s) (

) ) )

(s

comparing Eq.(1) with standard 2nd order equation

5.

[Ans. D] ( ) plot shows presence of 2 poles & 2 zeroes in Bode equivalent plot

6.

[Ans. D] m

√

s

ξ

3.

S.E(s)= 7.

[Ans. A]

(s )(s ) Only option (A) is satisfies.

s

t n

s (s )(s ) The lead compensator C(s) should first

(s) (s)

m s

(

)

for (D)

t n

for phase lead t n

term

(s (s

( ) ( ))

(

[Ans. A]

(s)

stabilize the plant i.e. remove

s

should be positive

t n

) Option (A) & (C) satisfies, it may be observed have be observed that option (C) will have poles and zero in RHS of s – plane, thus not possible (not a practical system) it can be concluded that option (A) is right

)

th

th

th


GATE QUESTION BANK

8.

[Ans. A] y ( )

t n

Control Systems

A lag lead compensator that provides an attention of 20dB and phase lead of 45 at the frequency of 3rad/sec is used.

t n

For maximum phase shift t

2.

[Ans. A] t n

t n

is always positive

t n

t n

is always negative

⁄

sl IN 1. √ r EE 1.

s

)(s

)

r

)|

|

. .

s

s(s

)(s

) (

)|

)

|

/ /

M

log

M

log (

| (

) ( )

(s (s

(s)

)

Negative GM implies that the system is unstable. should become gain cross over frequency, r s , the magnitude should be 0dB. o m t m gn tu ‘ ’ t r s a lag compensator which gives an attenuation of 20dB(before compensation the magnitude is 20dB) and to obtain 45 phase margin at r s a lead compensator with a phase lead of 45 is used.

)

(s) ) (s) s(s ol s o t omp ns t syst m r g v n s s s (s ) t (s) (s ) r t r st qu t on )(s ) (s ) s(s ) ,(s -(s ) ( )s ( )s s (s )(s ) s s s s ( )

√

) (

( (

|

(s)

( )

(s) (s) (s) (s)

√ | (

s l g omp ns tor

[Ans. C] Let the T.F of the compensator be o t omp ns t syst m

[Ans. D] s(s

n

2.

) )

[Ans. A] v n (s) {

m(s)

(

)} s

{ (s) m( ) s

th

* ( ) pl

th

(s)}

s

∑ ( )+

tr ns orm o

th

(t) t

I(s) s


GATE QUESTION BANK

3.

( ) 0 s For stable system all elements of first column should be greater than zero. By substituting all the given options in o nt o ’ w ll g v n g t v v lu except option D.

[Ans. B] The given diagram can be reduced as (s)

(s)

(s)

s

r ns r un t on (s)

( )

( )

( )

( )

Control Systems

6.

[Ans. C] Open loop transfer function of a dc motor as ( ) ( )

( )

R(s)

( )

(

(s)

(

(s)

)

d(t)= sin 0.2t or g tt ng mult ply ng |

tor Topic: P controller with unity feed back Formula: For first order system loop

|

|

|

√

(

( )

transfer function is

)

with

o mpl tu o output (t) n t l mpl tu

. Now for

( )

closed loop overall transfer function is given by .

[Ans. B] (PD controller can be used)

( )

/ .

/ (

5.

)

Dividing numerator and denominator by 10

[Ans. D] (s) (

)(

)

( s

)

(s)

ow

)( s

((s

)

)( s

( ) ( )

.

/

o

(s)

(s)

comparing

( )

( )

( )

4.

(s)

(s)

)

)

( y omp r ng rom ormul ) In Question given that time constant of closed loop system is

times of

( ) ( )

((

)(

)

)(

)

Characteristics equation is ( )s s s ( ) For stability, according to R-H array 6 (6+3k2) s ( 11 s ( ) 0 s

so .

)

/

approximate value

th

th

th


GATE QUESTION BANK

7.

Control Systems

[Ans. D] y(s)

(s)

(

(s)

y(s)) [

s

]

y(s) (s) (s) [

(s) [ s s

(s) [

]

] ]

Zero means at

th

th

th


GATE QUESTION BANK

Control Systems

State Variable Analysis ECE - 2006 1. A linear system is described by the following state equation ̇ (t)

̇ (t)

(t)

0

ECE - 2007 2. The state space representation of a separately excited DC servo motor dynamics is given as, +

0

10 1

0

The Eigen value and eigenvector pairs ( v ) for the system are

1

The state-transition matrix of the system is os t s nt (A) 0 1 s n t os t s nt ] (B) [ – os t s nt os t – os t s n t] (C) [ s nt os t os t s nt (D) 0 1 s nt os t

*

3.

4.

( )

(C)

(B)

(D)

x(0) = 0

1, then the system response is

x(t) = 0

1. If the initial state vector

of the system changes to x (0) = 0

0

1/

(B) .

0

1/ and .

0

1/

(C) .

0

1/ and . 0

1/

(D) .

0

1/ and . 0

1/

(B) 0

1 1

(C) 0

1

(D) 0

1

ECE - 2009 5.

Consider the system A=0

= Ax + Bu with

1 and B = 0 1 , where p and q

are arbitrary real numbers. Which of the following statements about the controllability of the system is true? (A) The system is completely state controllable for any nonzero values of p and q (B) Only P = 0 and q = 0 result in controllability (C) The system is uncontrollable for all values of p and q (D) We cannot conclude about controllability from the given data

1 u;

Common Data Questions 3 and 4 Consider a linear system whose state space representation is ̇ (t) = Ax (t). If the initial state vector of the system is

1/ and .

(A) 0

of the motor is

(A)

0

The system matrix A is

Where is the speed of the motor, is the armature current and u is the armature voltage. The transfer function ( )

(A) .

ECE - 2010 Common Data for Questions 6 and 7 The signal flow graph of a system is shown below.

1,

(s)

s

s

(s)

then the system response becomes x (t) = 0

1.

th

th

th


GATE QUESTION BANK

6.

The state variable representation of the system can be (A) (B) (C) (D)

7.

ẋ

0

1x

y

,

-x

ẋ

0

1x

y

,

-x

ẋ

0

1x

y

,

ẋ

0

y

,

0 1u 0 1u 0 1u

-x 1x

0 1u

-x

The transfer function of the system is (A)

(C)

(B)

(D)

ECE - 2011 8. The block diagram of a system with one input and two outputs 𝑦 and 𝑦 is given below. u

ECE - 2012 9. The state variable description of an LTI system is given by x ẋ x [ẋ ] [ ][ ] [ ]u x ẋ x - [x ] y , x Where y is the output and u is the input. The system is controllable for (A) a1≠ 2 =0, a3 ≠ (B) a1 =0, a2 ≠ 3≠ (C) a1=0, a2 ≠ 3 =0 (D) a1 ≠ 2 ≠ 3 =0 ECE - 2013 Statement for Linked Answer Questions 10 and 11: The state diagram of a system is shown below. A system is described by the state – variable equations ̇ u y u u

y

s

10. y

s

1

A state space model of the above system in terms of the state vector and the output vector 𝑦 ,𝑦 𝑦 - is , -x

(B) ẋ

,

(C) ẋ

0

(D) ẋ

0

, -u y

-x

, -u 1x

1x

, y

̇

0

y

,

̇

0 ̇

(C) ,

-x

0 1x 11.

,

̇

0

y

,

y

0

1u

0

1u

0

1u

0

1u

u 1 -

u 1

-

u 1

-

u

The state transition matrix of the system shown in the figure above is (A) 0 (B) 0

th

-

0

y (D)

1

,

y

-x

0 1u y

0 1u y

(B)

0 1x

1

1

1

1

The state – variable equation of the system shown in the figure above are (A)

(A) ẋ

Control Systems

th

(C) 0

1

t t

(D) 0

1

th

1 t

1


GATE QUESTION BANK

ECE - 2014 12. Consider the state space model of a system, as given below x ẋ x [ẋ ] [ ][ ] [ ]u x ẋ x - [x ] y , x The system is (A) controllable and observable (B) uncontrollable and observable (C) uncontrollable and unobservable (D) controllable and unobservable 13.

or x

s

s

x

x

[

1

(B) [

]

(C) [

]

(D) [ 16.

y

]

0 1 x(t) s

(A) 0

s

x

0 1 x(t) nx

Consider the state space system expressed by the signal flow diagram shown in the figure. u

Control Systems

]

The state transition matrix (t) of a x ẋ system [ ] 0 1 0x 1 is ẋ t (A) 0 (C) 0 1 1 t t (B) 0 (D) 0 1 1 t

EE - 2006 1. For a system with the transfer function (

H(s) =

)

, the matrix A in the

state space form ẋ = Ax + Bu is equal to The corresponding system is (A) always controllable (B) always observable (C) always stable (D) always unstable 14.

15.

An unforced linear time invariant (LTI) system is represented by x ẋ [ ] 0 1 0x 1 ẋ If the initial conditions are x (0)=1 and x ( ) t solut on o t st t equation is (A) x (t) x (t) (B) x (t) x (t) (C) x (t) x (t) (D) x (t) t x (t) The state equation of a second-order linear system is given by ẋ (t) x(t) x( ) x or x

0

1 x(t)

0

(A) [

]

(B) [

]

(C) [

]

(D) [

]

EE - 2008 Statement for Linked Answer Questions 2 and 3 The state space equation of a system is described by ẋ = Ax + Bu, y = Cx, where x is state vector, u is input, y is output 1 , B = 0 1, C = ,

and A = 0 2.

-

The transfer function G(s) of this system will be

(A)

1 n

(

(B) th

( th

(C)

)

(D)

) th

(

)

(

)


GATE QUESTION BANK

3.

A unity feedback is provided to the above system G(S) to make it a closed loop system as shown in figure.

r + ( t )

y ( t )

G(s)

Σ

For a unit step input r(t), the steady state error in the output will be (A) 0 (C) 2 (B) 1 (D) EE - 2009 Common Data Questions: 4 & 5 A system is described by the following state and output equations ()

x (t)

x (t)

()

x (t)

u(t)

4.

5.

EE - 2013 Common Data Questions 7 and 8 The state variable formulation of a system is given as x ẋ [ ] 0 1 0x 1 0 1 u x ( ) ẋ x - 0x 1 x ( ) n y , 7. The system is (A) Controllable but not observable (B) Not controllable but observable (C) Both controllable and observable (D) Both not controllable and not observable 8.

The response y(t) to a unit step input is (A) (B)

u(t)

y(t) = x1(t) Where u(t)is the input and y(t) is the output

(C) (D) EE - 2014 9. A system matrix is given as follows. [

The system transfer function is (A)

(C)

(B)

(D)

The state transition matrix of the above system is (A) [

]

(B) 0

1

(C) 0

1

(D) 0

0 (A) (B) (C) (D)

1

]

The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is _____ 10.

The second order dynamic system u

t

y the matrices P, Q and R as follows: 0

1

0 1

1

EE - 2010 6. The system ̇

Control Systems

, The system has the following controllability and observability properties: (A) Controllable and observable (B) Not controllable but observable (C) Controllable but not observable (D) Not controllable and not observable

with 0 1 is

stable and controllable stable but uncontrollable unstable but controllable unstable and uncontrollable th

th

th


GATE QUESTION BANK

11.

The state transition matrix for the system x ẋ [ ] 0 1 0x 1 0 1 u is ẋ (A) 0 (B) 0

(C) 0

1 t

1

(D) 0

t

Control Systems

IN - 2008 2. The state space representation of a ̇= 0

system is given by

1

y = [1 0]x. The transfer function

t 1

Consider the system described by following state space equations x ẋ [ ] 0 1 0x 1 0 1 u ẋ x - 0x 1 y , If u is unit step input, then the steady state error of the system is (A) 0 (C) 2/3 (B) 1/2 (D) 1

IN - 2006 1. The state-variable representation of a plant is given by ̇ = Ax + Bu , y = Cx. Where x is the state, u is the input and y is the output. Assuming zero initial conditions, the impulse response of the plant is given by (A) exp (At) (B) xp , (t - )] Bu () d (C) C exp (At) B (D) xp , (t - )] Bu () d

( ) ( )

of

the system will be (A)

12.

1 X + 0 1u,

(C)

(B)

(

(D)

)

IN - 2009 3. A linear time-invariant single-input single-output system has a state space model given by

=Fx + Gu; y = Hx

Where F =0

1; G=0 1; H =,

-.

Here, x is the state vector, u is the input, and y is the output. The damping ratio of the system is (A) 0.25 (C) 1 (B) 0.5 (D) 2 IN - 2011 4. The transfer function of the system described by the state-space equations x ẋ [ ] 0 1 0x 1 0 1 u ẋ x - 0x 1 is y , (A) (C) (B)

th

(D)

th

th


GATE QUESTION BANK

Control Systems


6. [Ans. A] ∅(t) = L *( I ( I

)

) +

0

 ( I

ẋ

)

) s nt 1 os t

0

 s

ẋ ẋ y

x u… … ( ) x x … … ( ) (x ẋ ) (x x x ) y x … … ( ) From equation 1, 2, 3 the state variable representation of the system is

(s)

I (s)

I (s) (s) u(s)

u(s)

( ) ( )

(

∅  ∅ ∅(t)

1x

𝑦

,

-

0 1u

[Ans. C] ,sI

∅ ∅ ] ∅ ∅ ∅ ∅

, &

∅ ∅

-0

, s

10 1

s (s

)

]

s s

) ) s

s s

-[

1 )

-

s

∅ ∅

0

(( I (s )(s find eigen vector

s s Method II It can also be solved by applying the M son’s g n ormul

&

[Ans. D] From above, , I 0

5.

0

[Ans. A] Let A = [

4.

ẋ

)

7. 3.

x ẋ

(s)

(s) ) (s)

(s) x

u (s) )(s

s

s

1

[Ans. A]

t  (s

ẋ

(s)

1

( os t (t) = 0 s nt

2.

[Ans. B]

-

0

8.

1

[Ans. B] (s) u(s) s (s) x (s) x (s) u(s) s x (s) y (s) u(s) s x (s) Similarly x (s) y (s) u(s) s u(s) so sx (s) x (s) u(s) y (s) x (s) sx (s) x (s) u(s)

1

[Ans. C] , 

-

p p 0q q1 p q  un ontroll

l

p, q

th

th

th


GATE QUESTION BANK

y (s) x (s) ẋ (t) x (t) u(t) y (t) ẋ (t) x (t) u(t) y (t) x ẋ [ ] 0 1 0x 1 0 1 u(t) ẋ x x u ,y y -

rom qu st on y 9.

x (t) x (t)

,

-

y=,

From the graph we can find that ,u x - ( ) ẋ ẋ x u… ( ) n ( ),x ẋ ẋ ẋ x ẋ From eq. (i) put value of ẋ ( x ẋ x u) ẋ x x u…… ( ) s m l rly y ẋ ( ) y x x u … ( ) ( ẋ x x ) ( u) [ ] ( ẋ x x ) u In matrix form x ẋ [ ] 0 1 0x 1 0 1 u ẋ x ẋ 0 1 0x 1 0 1 u And x , - 0x 1 u y -x u y ,

0 1

x ] [x ] + [ ]u x x - [x ] x

A=[

]& B = [ ]

So, AB = [ =*

][ ] = [ ] +

So, the controllability matrix, h ⌈

⌉ =[

]

11.

[Ans. A] ,s I

Determinant should be non-zero So, a1a2 (– a2) ≠ 1≠ a2 ≠ a3 may be 0 10.

y

u

[Ans. D] x *x + = [ x

Control Systems

0

-

s0

1

s -

0

sI

s

[Ans. A] For state variable form, we have to find number of integral (1/s) in the graph Assign output of these integral with state variable & input of these integral with derivative of state variable. 1. Then find out the relation between these derivative in terms of state variable and input 2. Similarly we can write the relation between output and these variable and input using state flow graph 3. Lets assume & as state variable

1

1

s

,sI

0

s s

1

(s

sI

)

) s [(s ] Take inverse Laplace transform both side ,sI

s

-

[(s 0

th

th

)

s

]

1

t

th


GATE QUESTION BANK

12.

[Ans. B] [

]

2

[ ]

, Controllable matrix ,

[

]

l

,

[Ans. B] Applying linearity property 0

o x(t) the state 16.

[Ans. D] (t)

0

0

1

[

, I

-

0

s

s s

EE 1.

x 1 0x 1

s ]

1

t

2

3 1t

0 0

1 1

t

3

(s) (s)

(s) ( )

( )

( )

( )

(s) (s) (s) s x t

… ) 0

1

[Ans. B]

x ( ) x ( )

t

t

0

(t)

0

1

s

The rank of S is 3 So the system is always controllable

t

1

s

+

(I

]

) -

[s

(t)

[Ans. C] ẋ [ ] 0 ⏟ ẋ

1

]

,( I s 0

The state controllable matrix is , ̇ ̇

14.

0 1

[

-[ ]

*

1

o

[Ans. A] From the state diagram, equations are ̇ [ ̇ ][ ][ ] [ ] ̇ 𝑦

15.

0 1

so o s rv

1

x (t) x ( ) x (t) x ( )

]

n

2 0

-

[

…

3

[

Rank is less than 3, so uncontrollable Observability matrix o , ( )

13.

Control Systems

(s

).

) s /

s s s (s) s x sx (s) u(s)

t

2

s

(s s

x (s)

Replacing s by x t

…

t

th

x t x t th

ẋ

x t

x

u(t) … ( )

x

th


3]

GATE QUESTION BANK

x ẋ x t Replacing eq.(i) ẋ x x x u(t) ẋ x x x u(t) ẋ x ẋ x ẋ x x x u(t) x ẋ [ẋ ] [ ] [x ] [ ] u(t) x ẋ o

2.

[

3.

,

- (0

s

5.

-

[Ans. B] 0 ( I

)

( I

)

0

s

(

m

(

s

( ))

(

(

0

1

x (t)

y(t)

,

x (t) -[ ] x (t)

y o

,

-&

, I

-

0

s s

s 0

s

s (s )(s r ns r un t on , I -

]

1

[Ans. C] , AB]  | ≠ controllable (s) ( I ) & (t) ( (s )) has exponential with positive power Unstable

7.

[Ans. A] , M

1 ontroll

,

M

8.

-

( I

)

, I

-

0

0

s

1

*, I

l

-

(s)+

1

s

1 )

l

ot o s rv

[Ans. A] Y(t) ,

1

s

s

6.

0

1

s

0

t(M ) ≠

0

]

s

)

t(M ) 1

)

s

)

0 1

y(t)

)(s

)

[Ans. C] Selecting x (t) and x (t) as state variables x (t) ẋ (t) x (t) x (t) u(t) t x (t) ẋ (t) x (t) u(t) t ẋ (t) x (t) [ ] 0 ] 0 1 u(t) 1[ ẋ (t) x (t) ẋ

, I

(s

[s m

0

1

s s

[Ans. A]

o

1

[

1 )0 1

s

s

1 s 0 1 (s )(s ) (s ) , -0 1 s (s )(s ) s s (s )(s ) s s ,

]

s 4.

0

[Ans. D] (s)

Control Systems

[s

] s

th

th

th


GATE QUESTION BANK

-

[, I x (t) [ ] x (t)

9.

)

[s(s

)]

,, I

s [(s * I

(s)-

-

[

12.

]

)(

(s)

0 ,

)

-

) )-

,( (

(s)

-

[Ans. C] 0

1

or ontroll ,

0 1

0

-

11.

1 2.

l

[Ans. C] 0

s

1+

1 s 0 1

s

)

s

rror

lm

(s)

l ms (

1

l m u(t) s

s(s

y(t) s

)

)

[Ans. A] (s) ( I

)

[

1 (

)

th

(sI

→

(

)

(s)

th

) s

)

) (s)

y(t)

 ( I

I

(sI

(t) → (sI

( I

State transition matrix s I 0 1 s s 0 s I (s )

0

s

s y st t

u(t) (s)

l

ot o s rv

1

[Ans. C]

1

0

s

t

(s) (s)

≠ ontroll For observability:

y(s) (s) )

(s)

s

s(s

l -

,

IN 1.

s

y(s) u(s)

v n

1

t

- *0 s

,

I

,

10.

0

[Ans. A]

|

,(

+

]

s

( I

[Ans. *] Range 2.9 to 3.1 Characteristic equation |

)

r ns r un t on

x (t) -[ ] x (t)

,

Y(t)

s(s

(s)]

Control Systems

)

]

(

(

)

.,

-0

)

0

1

-0

., 1/

th

)

1 0 1/ (s)


GATE QUESTION BANK

3.

Control Systems

[Ans. B] The poles of the s/y are obtained from the s/y matrix, f det (SI – F )= 0 0

1

S(S + 2) + 4 = 0

4.

[Ans. A] T(s) = ( I , (

) -

(

)

0

10 1

)

th

th

th


GATE QUESTION BANK

Analog Circuits

Diode Circuits - Analysis and Application ECE - 2006 1. For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 Volts. The waveform observed across R is

ECE - 2007 2. The correct full wave rectifier circuit is (A)

Input Output

~

i

(B)

(A)

6V

Input

6V

(B)

Output

(C)

12V

(C)

12V Input Output

6V (D)

(D)

Input

6V

Output

3.

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 . If the input voltage ( ) range is from 10 to 16 V, the th

th

th


GATE QUESTION BANK

Analog Circuits

output voltage ( ) range from

. The small signal input where

.

i

(A) 7.00 to 7.29V (B) 7.14 to 7.29 V

~

(C) 7.14 to 7.43 V (D) 7.29 to 7.43 V

ECE - 2008 4. In the following limiter circuit, an input voltage = 10sin100 t is applied. Assume that the diode drop is 0.7V when it is forward biased. The Zener breakdown voltage is 6.8V.The maximum and minimum values of the output voltage respectively are

6.

The bias current (A) 1 mA (B) 1.28 mA

7.

The ac output voltage (A) (B) (C) (D)

8.

A Zener diode, when used in voltage stabilization circuits, is biased in (A) reverse bias region below the breakdown voltage (B) reverse breakdown region (C) forward bias region (D) forward bias constant current mode

1k D1

through the diodes is (C) 1.5 mA (D) 2 mA is

D2 Z

(A) 6.1V, 0.7 V (B) 0.7 V, 7.5 V

6.8 V

(C) 7.5 V, 0.7 V (D) 7.5 V, 7.5 V

ECE - 2009 5. In the circuit below, the diode is ideal. The voltage V is given by

(A) min ( (B) max (

) )

(C) min ( (D) max(

ECE - 2012 9. The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode is

~

) )

ECE - 2011 Statements for Linked Answer Questions 6 and 7 In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7V. The thermal voltage

(A) (B) i

th

th

–1

(C) 1 – (D) 1 – i

th


GATE QUESTION BANK

ECE/EE/IN - 2012 10. The I-V characteristics of the diode in the circuit given below are i

(A) (B) (C) (D)

sin (sin (sin 0 for all t

Analog Circuits

i i

)/2 )/2

{ ECE - 2014 13. In the figure, assume that the forward voltage drops of the PN diode and Schottky diode are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF denotes nonconducting state of the diode, then in the circuit,

i

The current in the circuit is (A) 10 mA (C) 6.67 mA (B) 9.3 mA (D) 6.2 mA ECE/EE/IN - 2013 11. In the circuit shown below, the knee current of the ideal Zener diode is 10 mA. To maintain 5 V across RL, the minimum value of RL in  and the minimum power rating of the Zener diode in mW, respectively , are

(A) both and are ON (B) is ON and is OFF (C) both and are OFF (D) is OFF and is ON 14.

100

ILoad 10V

RL

VZ=5V

(A) 125 and 125 (B) 125 and 250 12.

The diode in the circuit shown has = 0.7 Volts but is ideal otherwise. If = 5sin(𝜔𝑡)Volts, the minimum and maximum values of (in Volts) are, respectively,

(C) 250 and 125 (D) 250 and 250

A voltage 1000 sin Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts, is

(A) 5 and 2.7 (B) 2.7 and 5

(C) 5 and 3.85 (D) 1.3 and 5

1k W

Y Z

15. X

1k

th

The figure shows a half-wave rectifier. The diode D is ideal. The average steadystate current (in Amperes) through the diode is approximately __________

th

th


GATE QUESTION BANK

+

Analog Circuits

+

D1 D2

i z

~

Vi

V0

RL =

5V

10V

-

16.

(A)

Two silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the figure. The range of input voltage for which the output voltage is

10

10

(B)

(A) (B) (C) (D)

5 5

EE - 2006 1. What are the states of the three ideal diodes of the circuit shown in figure?

(C) 10 5 5 10

10V

5A

(D) (A) (B) (C) (D)

ON, OFF, ON, OFF,

OFF, ON, OFF, ON,

OFF OFF ON ON

10

10 2.

Assuming the diodes D1 and D2 of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be th

th

th


GATE QUESTION BANK

Analog Circuits

(B)

EE - 2007 3. The three – terminal linear voltage regula r i e e a l a resistor as shown in the figure. If Vin is 10 V, what is the power dissipated in the transistor? +10V

(C) RL Vin 6.6V Zener diode 0

(A) 0.6 W (B) 2.4 W

(D)

(C) 4.2 W (D) 5.4 W

EE - 2008 4. The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. +

+

0.7 V

5.

In the voltage doubler circuit shown in he igure he wi h ‘S’ i l e a Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero, the steady state voltage across capacitors C1 and C2 will be

+

~

𝜔𝑡 S

~ If such diodes is used in clipper circuit of figure given above, the output voltage (v0) of the circuit will be

i

(A) (B) (C) (D)

(A)

= 10V, = 10V, = 5V, = 5V,

= 5V = 5V = 10V = 10V

EE - 2009 6. The following circuit shown has a source voltage Vs as shown in the graph. The current through the circuit is also shown th

th

th


GATE QUESTION BANK

Analog Circuits

a

10V

15V

15 10

(A) 4V (B) 5V

Vs (Volts)

5 0

EE - 2011 8. A clipper circuit is shown below.

-5

1k

-10 -15

0

100

200 300 Time (ms)

400

D

~

1.5 Current (mA)

(C) 7.5V (D) 12.12V

5V

1.0 0.5

Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is

0 -0.5 -1.0 -1.5

(A) 0

100

200 300 Time (ms)

400

The element connected between a and b could be

4.3

a

4.3 a

a

(B) 10

a

4.3

EE - 2010 7. Assuming that the diodes in the given circuit are ideal, the voltage is

4.3

th

th

th

10


GATE QUESTION BANK

Analog Circuits

dynamic resistance of the diode at room temperature is approximately,

(C)

5.7

1.7V 0.7 0.7 5.7

(A) (B) (D) 10

10

(C) (D)

IN - 2011 2. Assuming zener diode has currentvoltage characteristics as shown below on the right and forward voltage drop of diode is 0.7 V, the voltage in the circuit shown below is

EE - 2014 9. The sinusoidal ac source in the figure has an rms value of

√

. Considering all

possible values of , the minimum value of in to avoid burnout of the Zener diode is __________

I

2.7V

√

V

0.7V

~ (A) 3.7 V (B) 2.7 V

10.

Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage must be outside the range

~ (A) (B)

(C) 2.2 V (D) 0 V

IN - 2014 3. For the circuit shown in the figure assume ideal diodes with zero forward resistance and zero forward voltage drop. The current through the diode in mA is ___________.

(C) (D)

IN - 2008 1. In the circuit shown below, the ideality factor of the diode is unity and the voltage drop across it is 0.7V. The th

th

th


GATE QUESTION BANK

4.

Analog Circuits

For the circuit shown in the figure, the ra i r ha β =0.7 V, and the voltage across the Zener diode is 15 V.The current (in mA) through the Zener diode is ___________.


[Ans. B] When , zener diode is in forward bias so When , zener diode is in reverse bias, so when , zener diode will be OFF and . When , zener diode will be in breakdown region and .

2.

[Ans. C] Option – C, Circuit makes Full –wave rectifier.

3.

[Ans. C] For zener

When

4.

[Ans. C] When , is forward bias and are OFF, and When is +10V, are ON and zener diode is in reverse bias so V

5.

[Ans. A] When , diode is ON and V = When , diode is OFF and V = 1 V So V = min (

6.

[Ans. A]

7.

[Ans. B]

and

r

r

When

, current

~ th

th

r

th


GATE QUESTION BANK

10.

Analog Circuits

[Ans. D]

i

8.

[Ans. B] Zener diode operated in reverse in breakdown region.

i i r Since diode will be forward biased voltage across diode will be 0.7 V i

Stabilized

9.

11.

[Ans. A]

[Ans. B] i

~

5v

40 mA

10 V

la per

ea

e i ier

10 mA

he ex i e y he la pi g section clamp the positive peak to 0 volts and negative peak to 2 volts. So whole cos ( i l wer y 1 volts

When zener starts will be held @ 5V. i

pu

When zener starts the current drawn from supply is 50 mA at 10 V p wer drawn = 500 mW This should be dissipates in circuit power i ipa e i Resistance = Remaining 250 mW has to be dissipated by zener assuming worst bad ze er h ul e ra e r

th

th

th


GATE QUESTION BANK

12.

[Ans. D]

14.

Analog Circuits

[Ans. C]

1k y i tz

i

1k w

X

When Z is relatively positive to y Both D1& D2 are & l a Gets shorted.

When

positive maximum

i

When y is relatively positive to Z all Diode Biased urre ll w always. 13.

[Ans. D]

l

When

negative minimum

i

Assume both the diode ON. Then circuit will be as per figure (2)

15.

w p is off and hence

i le

i

[Ans. *] Range 0.08 to 0.12 For average steady state case, capacitor is open circuited S i

th

a pere

th

th


GATE QUESTION BANK

16.

[Ans. D]

3.

Analog Circuits

[Ans. B] So current through Current through 1k ,

i

So

mA

mA

= 0.5893 Power dissipated in transistor i

i

r hi W

4.

i

[Ans. A] When diode is OFF, the equivalent circuit is shown as follows: i

i

EE 1.

i

r hi

[Ans. A] Let is ON, OFF and equivalent circuit

10 V

i

~ i

OFF, then

i.e. when is max i.e., so diode never conducts and it is always OFF. So, i

5A

In this case,

5.

10A

[Ans. D] When , is ON and charges upto 5V and When , is ON and charges by -10 V and So in steady state, and

A So voltage across diode So is in reverse bias i.e. it is OFF. Voltage across diode V it is also in Reverse bias so OFF. Voltage across diode it is in forward bias and ON.

2.

[Ans. A] When

, ,

and

is ON and

6.

is OFF so . is OFF, so

[Ans. A] It is

are OFF, so When

OFF so,

mA, and diode conducts , diode will be reverse bias and I will become zero.

th

th

th


GATE QUESTION BANK

7.

Analog Circuits

Hence for output to be clipped input should not be lie inside this range

[Ans. B] Right side diode is off So ,

8.

9.

[Ans. C] When output will follow input, because zener diode and normal diodes are off When V Zener diode forward bias and V When V Diode is forward bias and

IN 1.

[Ans. B] Dynamic resistance r

r 2.

[Ans. C] First Assume that the diode does not reach reverse breakdown. So circuit becomes


i

10.

[Ans. B]

S S This is less than reverse breakdown voltage of diode . So our assumption is correct

i

i

3.

[Ans. 10] Diode goes to forward biased due to 10 V and 8V across diode and current through diode is

4.


i

i

Given circuit is union of both the above circuit for range

th

th

th


GATE QUESTION BANK

Analog Circuits

β

Nodal at

th

th

th


GATE QUESTION BANK

Analog Circuits

AC & DC Biasing-BJT and FET ECE - 2006 1. An n- channel depletion MOSFET has following two points on its curve: (i) d (ii) Which of the following Q-points will give the highest trans-conductance gain for small signals. (A) (B) (C) (D) ECE - 2007 2. The DC current gain ( ) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is (A) 0.980 (C) 0.990 (B) 0.985 (D) 0.995 3.

For the BJT circuit shown, assume that the of the transistor is very large and = 0.7 V. The mode of operation of the BJT is

The value of current is approximately (A) 0.5 mA (C) 9.3 mA (B) 2 mA (D) 15 mA ECE - 2011 5. For a BJT, the common – base current g i α .98 d he c ec r b e junction reverse bias saturation current . . This BJT is connected in the common emitter mode and operated in the active region with a base drive current . The collector current for this mode of operation is (A) 0.98 mA (C) 1.0 mA (B) 0.99 mA (D) 1.01 mA 6.

10 kΩ

2V

9.

In the circuit shown below, for the MOS transistor, and the threshold voltage . The voltage at the source of the upper transistor is 6V

1 kΩ

5V

(A) cut-off (B) saturation

W/L = 4

(C) normal active (D) reverse active W/L = 1

ECE - 2010 4. In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2.

(A) 1 V (B) 2 V

th

th

(C) 3 V (D) 3.67 V

th


GATE QUESTION BANK

ECE - 2013 7. In the circuit shown below, the silicon npn transistor Q has a very high value of . The required value of R2 in k to produce IC = 1mA is

10.

Analog Circuits

For the MOSFETs shown in the figure, the threshold voltage | | = 2 V and ( )

.

. The value of

(in mA) is ______.

VCC 3V

R1 60 k

IC

Q R2

(A) 20 (B) 30 8.

RE 500 

(C) 40 (D) 50

11.

In a MOSFET operating in the saturation region, the channel length modulation effect causes (A) an increase in the gate – source capacitance (B) a decrease in the transconductance (C) a decrease in the unity –gain cutoff frequency (D) a decrease in the output resistance

ECE - 2014 9. For the n-channel MOS transistor shown in the figure, the threshold voltage is 0.8 V. Neglect channel length modulation effects. When the drain voltage =1.6 V, the drain current was found to be 0.5 mA. If is adjusted to be 2 V by changing the values of R and , the new value of (in mA) is

For the MOSFET shown in assume W/L = 2, = 2.0 V, 100 / and = transistor switches from region to linear region when is__________

the figure, 0.5 V. The saturation (in Volts)

EE - 2006 1. Consider the circuit shown in figure. If the of the transistor is 30 and ICBO is 20 nA and the input voltage is + 5 V, then transistor would be operating in +12V . Ω

Ω

Q Ω

(A) 0.625 (B) 0.75

(C) 1.125 (D) 1.5

12V

(A) saturation region (B) active region th

th

th


GATE QUESTION BANK

Analog Circuits

(C) breakdown region (D) cut – off region 2.

A TTL NOT gate circuit is shown in fig. Assuming . of both the transistor, if . then the states of the two transistors will be, +5V

(A) 0 mA (B) 3.6 mA

.

(C) 4.3 mA (D) 5.7 mA

EE - 2010 5. The transistor circuit shown uses a silicon transistor with = 0.7V, and a dc current gain of 100. The value of is +10 V

(A) (B) (C) (D)

d e er e e er e d

d d e er e

50 k

10 k

EE - 2007 3. The common emitter forward current g i f he r i r h w i F = 100.

100 

+10V Ω

Ω

(A) 4.65V (B) 5V 6.

(C) 6.3V (D) 7.23V

Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal , the I-V characteristic of the composite switch is

Ω

The transistor is operating in (A) Saturation region (B) Cutoff region (C) Reverse active region (D) Forward active region

V

+

I

EE - 2008 4. Two perfectly matched silicon transistors are connected as shown in the figure. Assuming the of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is

I

(B)

(A)

V

V

th

th

I

th


GATE QUESTION BANK I

I

(D)

(C) V

V

IN - 2006 1. The biasing circuit of a silicon transistor is shown below. If  = 80, then what is VCE for the transistor? +12V

EE - 2011 7. The transistor used in the circuit shown be w h f d is negligible. 15k 1k

Analog Circuits

RB

RC

Ω

2.2k

Ω

8

D = 0.7V .

(A) 6.08 V (B) 0.2 V

(C) 1.2 V (D) 6.08 V

IN - 2007 2. In the circuit shown below,

.

.

If the forward voltage drop of diode is 0.7V, then the current through collector will be (A) 168 mA (C) 20.54 mA (B) 108 mA (D) 5.36 mA EE - 2014 8. The transistor in the given circuit should always be in active region. Take = 0.2 V, = 0.7 V. The maximum value of in . Which can be used, is__________

.

The f he r respectively, (A) 19 and 2.8 V (B) 19 and 4.7 V (C) 38 and 2.8 V (D) 38 and 4.7 V 3.

i

r

d

are,

The three transistors in the circuit shown below are identical, with = 0.7 V and .

th

th

th


GATE QUESTION BANK

Analog Circuits

IN - 2014 6. In the figure, transistors T and T have identical characteristics. of transistor T is 0.1 V. The voltage is high enough to put T in saturation. Voltage of transistors T T and T is 0.7 V. The value of ( ) in V is ___________.

10 V Ω 2V Ω

9

The voltage (A) 0.2 V (B) 2 V

T

is (C) 7.4 V (D) 10 V

IN - 2010 4. The matched transistors and shown in the adjoining figure have =100. Assuming the base-emitter voltages to be 0.7V, the collector-emitter voltage V2 of the transistor is

(A) 33.9V (B) 27.8V

T

T

(C) 16.2V (D) 0.7V

IN - 2013 5. In the transistor circuit as shown below, the value of resistance RE in k is approximately, +10V

1.5k 15k .

(A) 1.0 (B) 1.5

6k

IC=2.0 mA . VCE=5.0 V RE

Vout

(C) 2.0 (D) 2.5

th

th

th


GATE QUESTION BANK

Analog Circuits


5.

[Ans. D] α

[Ans. D] From the below graph it is clear that a increase conductance i.e. slope of graph is increased.

α .98 .98 . 6.

9 .

[Ans. C] The transistor which has

12mA

and

6 V

So that transistor in saturation region. The transistor which has

2.

Transfer characteristic of n-channel DMOSFET.

Drain is connected to gate So that transistor in saturation

[Ans. B]

The current flow in both the transistor is same

α

( ) (

α base transport factor × emitter injection efficiency b e r

r f c r

( ) (

)

.99 .98

3.

)

[Ans. B] Given is large so d Assuming BJT is in active Applying KVL in Base. Emitter loop . . . 9.8 w .9

8

7.

[Ans. C]

.

So BJT is in saturation 4.

[Ans. B] Assuming . .

~

d

.

i ce is very large

=1mA

So,

eg igib e

So, = 2.04 mA

ge dr

2 mA th

th

cr th

e i er f

re i


r

GATE QUESTION BANK

11. . w

Analog Circuits

[Ans. *] Range 1.4 to 1.6 Initially the transistor is in saturation [

. .

.

( )]

.

.

w

r

gi e circ i

r .

.

r .

b i

.

i g

i

. .

. .

.8 .

8.

[Ans. D]

9.

[Ans. C] ere

EE 1. i

r i

.

[Ans. B] Let the device is in saturation and . V

regi

.

Collector current .

.

.8

For saturation

.

Base voltage . current through 100

. .

.8

. .

. .

10.

. .

= 5.3mA

= 0.178 mA re i

ce

.

= 0.127 mA c rre hr .

.

[Ans. *] Range 0.88 to 0.92

gh

re i

ce

= 0.29 mA Base current

. 9 . = 0.163 mA So this is less than it means device is not in saturation. It is in active region.

NMOS is in saturation region . . .9

2.

[Ans. C] When . he will be in reverse active mode .i.e. Reverse on and will be ON.

3.

[Ans. D] We assume BJT is in active region applying KVL in base emitter circuit . i g th

th

th


GATE QUESTION BANK

8.

[Ans. *] Range 22 to 23 To calculate max, take end condition . i . . i

9. 9 .

Ti i 4.

Analog Circuits

.9

i

ci e c i e regi

. .

. .8

[Ans. B]

.

. IN 1.

[Ans. B] .

. 9 9 . and drop across diode = 0.7 . . . i

So device is not active region. Let the device is in saturation and . V . .9

ery rge

As both transistors are perfectly matched and . of both transistors. . mA 5.

.9 8

. .

[Ans. A]

So device is in saturation, so

.

. 9.

2.

. .

[Ans. A]

.

6.

[Ans. C] When V will be + ve both transistor and diode will be on making V across them Zero and Current I will be flow and when V is -ve both will be off offering infinite resistance so current I will be Zero.

7.

[Ans. D] Transistor is in Saturation region . . .

.

. . th

th

th


GATE QUESTION BANK

Apply KVL .

Analog Circuits

. 8 For matched transistor .

.

di is same for both the transistors) Now apply KVL in the outer port of transistor . 8 .8

9 9. Apply KVL to output terminals 5. .9 3.

9. .

[Ans. A] 10 V

. .8

2m A

.

Ω

V=IR=1.5k × 2mA=3V

15k

[Ans. C]

=10-3=7V

10V

gi e

2V 6k

gi e

Using KVL in the left hand transistor, .

Ω

be .

6.

gi e

[Ans. *] Range 5.5 to 5.8 9

Now again apply KVL to the collector resistor of the right and transistor T .

. .

.

. 4.

[Ans. B] rre hr .

T

T .

gh

re i

.

ce

.

i

Apply KVL on path 1 9 . .

. .

=1.66mA

. th

th

th


GATE QUESTION BANK

Tr

i

. r ide ic

y

h .

Analog Circuits

– . . .98

. .

th

th

th


GATE QUESTION BANK

Analog Circuits

Small Signal Modeling of BJT and FET ECE - 2006 Common Data Questions 1, 2, 3 In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters: h h The capacitance can be assumed to be infinite.

defined to be

2.

3.

In the figure above, the ground has been shown by the symbol Under the DC conditions, the collector-toemitter voltage drop is (A) 4.8 Volts (C) 6.0Volts (B) 5.3 Volts (D) 6.6Volts

ECE - 2008 4. Two identical NMOS transistors M1 and M2 are connected as shown below. is chosen so that both transistors are in saturation.The equivalent g of the Pair is

of the of the

Statement for Linked Answer Questions 5 and 6 In the following transistor circuit,

If is increased by 10%, then collectorto-emitter voltage drop (A) increases by less than or equal to 10% (B) decreases by less than or equal to 10% (C) increases by more than 10% (D) decreases by more than 10% The small-signal gain of the amplifier ⁄ is (A) 10 (C) 5.3 (B) 5.3 (D) 10

.

The equivalent g of the pair is (A) the sum of individual g transistors (B) the product of individual g transistors (C) nearly equal to the g of M1 (D) nearly equal to g ⁄g of M2

~

1.

at constant

, and

and all the

capacitances are very large.

5.

The value of DC current is (A) 1mA (C) 5mA (B) 2mA (D) 10mA

6.

The mid-band voltage gain amplifier is approximately (A) 180 (C) 90 (B) 120 (D) 60 th

th

th

of

the


GATE QUESTION BANK

ECE - 2009 7. A small signal source () A +B i is applied to a transistor amplifier as shown below. The transistor has =150 and h =3k .Which expression best approximates ( )?

(B) The input resistance decreases and the magnitude of voltage gain increases (C) Both input resistance and the magnitude of voltage gain decrease (D) Both input resistance and the magnitude of voltage gain increase

12V

3k

100k

V0(t) 100 nF Vi(t) 100nF 20k

(A) (B) (C) (D)

( ( ( (

)= )= )= )=

μF

900k 

ECE - 2012 9. The current ib through the base of a silicon npn transistor is 1 + 0.1 cos(10000 t) mA . At 300 K, the r in the small signal model of the transistor is i

(

)

i

(

Analog Circuits

i

)

i i

ECE - 2010 8. The amplifier circuit shown below uses a silicon transistor. The capacitors and can be assumed to be short at signal frequency and the effect of output resistance can be ignored. If is disconnected from the circuit, which one of the following statements is TRUE?

(A) (B)

Ω Ω

(C) (D)

Ω Ω

ECE - 2013 10. The small-signal resistance(i.e., dVB/dID) in k offered by the n – channel MOSFET M shown in the figure below, at a bias point of VB =2V is (device data for M: device transconductance parameter 2 kn μnCox(W/L) 4 μ / , threshold voltage VTN=1V and neglect body effect and channel length modulation effects) ID

VB

~

M

(A) The input resistance increases and the magnitude of voltage gain decreases

(A) 12.5 (B) 25 th

th

(C) 50 (D) 100 th


GATE QUESTION BANK

Analog Circuits

ECE - 2014 11. For the amplifier shown in the figure, the BJT parameters are and thermal voltage = 25 mV. The voltage gain ( / ) of the amplifier is _____ .

(A) g (B) μF

μF

F

12.

(C) g (D)

14.

A BJT in a common-base configuration is used to amplify a signal received by a 50 antenna. Assume kT/q = 25 mV. The value of the collector bias current (in mA) required to match the input impedance of the amplifier to the impedance of the antenna is __________

15.

For the common collector amplifier h w i he figu e he JT h high negligible = 0.7 V. The ( ) and maximum undistorted peak-to-peak output voltage (in Volts) is

In the circuit shown, the PNP transistor has = 0.7 Vand = 50. Assume that = 100 k . For to be 5 V, the value of (in k .) is __________

μF μF

13.

Consider the common-collector amplifier in the figure (bias circuitry ensures that the transistor operates in forward active region, but has been omitted for simplicity). Let be the collector current, be the base-emitter voltage and be the thermal voltage. Also, g are the small-signal transconductance and output resistance of the transistor, respectively. Which one of the following conditions ensures a nearly constant small signal voltage gain for a wide range of values of ? th

th

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GATE QUESTION BANK

ECE/EE/IN - 2012 1. The voltage gain below is

(A) 0.967 (B) 0.976

of the circuit shown

~

(A) (B)

(C) (D)

EE - 2014 2. The magnitude of the mid-band voltage gain of the circuit shown in figure is (assuming h of the transistor to be 100)

Analog Circuits

(C) 0.983 (D) 0.998

IN - 2008 2. For a single stage BJT common base amplifier (A) current gain as well as voltage gain can be greater than unity (B) current gain can be greater than unity but voltage gain is always less than unity (C) voltage gain can be greater than unity but current gain is always less than unity (D) current gain as well as voltage gain is always less than unity 3.

In the amplifier circuit shown below, assume VBE he f he transistor and the values of C1 and C2 are extremely high. If the amplifier is designed such that at the quiescent point

h

its VCE =

~

where VCC is the power supply

voltage, its small signal voltage gain |

|

will be

(A) 1 (B) 10

(C) 20 (D) 100

IN - 2006 1. An amplifier circuit is shown below. Assume that the transistor works in active region. The low frequency smallsignal parameters for the transistor are g = 20 mS, 0 = 50, =, =0. What is the voltage gain,

( ) of the

amplifier? (A) 3.75 (B) 4.5

+Vcc

Ω

Vi

~

Ω

(C) 9 (D) 19

IN - 2009 Common Data Questions 4 and 5 The circuit shown in the figure uses three identical transistors with = 0.7V and Gi e : = = Ω th

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GATE QUESTION BANK

kT/ = 25mV. The collector current of transistor 2mA.

4.

The bias voltage at the base of the transistor is approximately (A) 9.3V (C) 10.3V (B) 10.0V (D) 11.0V The small signal voltage gain of the circuit is (A) 20 (C) 20 (B) 40 (D) 40

is

+12V

5. R1

R2

Q1

Analog Circuits

Q2 Q3 R3 12V


So, [Ans. C] Under DC conditions capacitor as open. By KVL,

)

g

will act

( ) Which is equal to the g of

but μ

(

.

[Ans. A] Equivalent circuit

5.

= 6V 2.

[Ans. B] in increased by 10%, i.e. By KVL,

4μ i e ge By KVL, in E – B loop, e e e 3.

[Ans. A]

4.

[Ans. C] Both transistor carry same current and both are in saturation

[Ans. D]

6.

Voltage gain u

th

i

th

e

(

)(

)

ge

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GATE QUESTION BANK

11.

Analog Circuits

[Ans. *] Range – 240 to – 230 From the model i

i

(

7.

[Ans. B] The e is

i

e

we f

u u

) e

i

ge

h h (

)

i

For the calculation of gain we need to calculate . so DC analysis is required

8.

[Ans. A] The moment is disconnected from the circuit, h ( h ) but with capacitor, h which also reduces voltage gain So, increases and decreases.

9.

[Ans. C]

g

i = 1 + 0.1 cos ( Calculate Solution: [

) = 3V

whe e

i he he

ge

e u e

44

]

4

10.

[Ans. B] μ L

(

) μ

(

L μ 4 4

L

) i

(

) (

i

)

(

i

4 th

th

th

)i


GATE QUESTION BANK

Analog Circuits

i i i i

μ

g

i

e

i i

i

e

4 4 he e e figu e i i ( ) i i ( )i ( ) ) ( i ( ) i ( ) i

f

4 4

e

Now assuming that collector current g emitter terminal. g

is very large, the flows through the

g

4 4

g g

12.

So if g constant g

[Ans. *] Range 1.04 to 1.12 (

)

(

)

which is nearly

14.

[Ans. *] Range 0.49 to 0.51 For CB configuration

15.

[Ans. *] Range 9.39 to 9.41

4 4 13.

[Ans. B] This is a common-collector configuration.

i e eg i e can vary ( )i between 12V in the positive direction and 0V in the negative direction biased around So, positive excursion of output voltage = 4 Negative excursion of output voltage

i u

` Replacing the transistor equivalent model, we get

with

T

–

So, undistorted peak to peak output voltage 4 4 th

th

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GATE QUESTION BANK

Analog Circuits

Apply miller’s theorem ge i

u e i i

i e eg i e can vary ( )i between 12V in the positive direction and 0V in the negative direction biased around So, positive excursion of output voltage = 4 Negative excursion of output voltage

i

i ( ( ) i i ( ( ) i

[Ans. D] In DC Analysis capacitor will behave as an open circuit

i

(

)

whe e

So, undistorted peak to peak output voltage 4 4 EE 1.

i

~

4

) ) 4

(

(

)

4

Consider input section

i

~

i i

4

i

(i i

i )

i

)i

i

i (

i

(

i i i

i (

(

)

4

(

)

) 2.

)i

[Ans. D] g

whe e

g Small signal analysis of circuit

i g and Q point to be in the middle of the load line

i

~

(

)

i

g

th

th

th


)

GATE QUESTION BANK

IN 1.

Analog Circuits

By Voltage division rule [Ans. D] Apply KVL to the base emitter circuit

Ω ~

Ω

Vi

w

AC equivalent circuit for

L

he

e

e i e

i ui

model

i

i

4 g

G i

~

|

|

4 | |

①

4.

[Ans. A] Emitter

voltage

of

transistor,

g L i Where i i i

10 V

① i i

(i i (i (

i

g

)

So i )

g

5.

) i ( (i [ 2.

3.

[Ans. B] When output is taken between one collector and ground then

)i )

g i i i

g

(i e

i g

e

e g i )

and if take the output between two collectors then differential gain. g collector resistance

]

[Ans. C] For CB, current gain is close to unity while voltage gain is very high.

g

/ and here output is taken between one collector and ground therefore gain g

[Ans. A]

4

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GATE QUESTION BANK

Analog Circuits

BJT and JFET Frequency Response ECE - 2010 Common Data for Questions 1 and 2 Consider the common emitter amplifier shown below with the following circuit parameters: g A/V, ,

RD 10 k

C V0

Vi

M RL 10 k

and (A) 8 (B) 32

(C) 50 (D) 200

ECE - 2014 4.

1.

The resistance seen by the source v is (A) (C) (B) (D)

2.

The lower cut off frequency due to (A) 33.9 Hz (C) 13.6 Hz (B) 27.1 Hz (D) 16.9 Hz

A cascade connection of two voltage amplifiers and is shown in the figure. The open-loop gain , input resistance , and output resistance for and are as follows:

The approximate overall voltage gain / is __________. is

ECE - 2013 3. The ac schematic of an NMOS common – source stage is shown in the figure below, where part of the biasing circuits has been omitted for simplicity. For the nchannel MOSFET M, the transconductance gm =1mA/V, and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in Hz of the circuit is approximately at

IN - 2011 1. The amplifier shown below has a voltage gain of an input esistance of and a lower 3-dB cut-off frequency of 20 Hz. Which one of the following statements is TRUE when the emitter resistance is doubled?

th

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GATE QUESTION BANK

Analog Circuits

v

v

~

(A) Magnitude of voltage gain will decrease (B) Input resistance will decrease (C) Collector bias current will increase (D) Lower 3-dB cut-off frequency will increase

th

th

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GATE QUESTION BANK

Analog Circuits


[Ans. B] Equivalent model of the given circuit is shown below

h

~

[ IN 1.

h

]

[Ans. A] The small signal model is

g The resistance seen by the source h

h

2.

h

[Ans. B] Lower cut-off frequency due to

h

f h increases

3.

c

w

ec eases ec ease

f f 4.

doubled Zi

ou e ec eases owe frequency decreases as increases s inc eases

[Ans. A] w

h

ec eases

≅8

[Ans. *] Range 34 to 34.72

p ifie

p ifie

th

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GATE QUESTION BANK

Analog Circuits

Feedback and Oscillator Circuits ECE - 2006 1. The input impedance ( ) and the output ( ) impedance of an ideal transconductance (voltage controlled current source) amplifier are (A) (C) (B) (D)

ECE - 2011 4. In the circuit shown below, capacitors and are very large and are shorts at the input frequency. is a small signal input. The gain magnitude | ⁄ | at 10 Mrad/s is 5V 2k𝛀

ECE - 2007 2. In a transconductance amplifier, it is desirable to have (A) A large input resistance and a large output resistance (B) A large input resistance and a small output resistance (C) A small input resistance and a large output resistance (D) A small input resistance and a small output resistance ECE - 2009 3. In the circuit shown below, the op-amp is ideal the transistor has Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp

1nF

μH

2k𝛀

2.7V 2k𝛀 ~

(A) Maximum (B) Minimum

(C) Unity (D) Zero

ECE - 2014 5. In the ac equivalent circuit shown in the figure, if is the input current and is very large, the type of feedback is

10V

V

(A) (B) (C) (D)

+

(A) voltage-voltage feedback (B) voltage-current feedback (C) current-voltage feedback (D) current-current feedback

Positive Feedback V=10V Positive Feedback V= 0V Positive Feedback V=5V Positive Feedback V=2V th

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GATE QUESTION BANK

6.

The feedback topology in the amplifier circuit ( the base bias circuit is not shown for simplicity) in the figure is

Analog Circuits

(A) 5

10 (B) 10 5

~ (C)

(A) Voltage shunt feedback (B) Current series feedback (C) Current shunt feedback (D) Voltage series feedback 7.

5

10

The desirable characteristics of a transconductance amplifier are (A) high input resistance and high output resistance (B) high input resistance and low output resistance (C) low input resistance and high output resistance (D) low input resistance and low output resistance

EE - 2006 1. A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are 12V. The voltage waveform at point P will be

(D)

10

5

EE - 2009 2. The nature of feedback in the op-amp circuit shown is +6V 2

K

R1 Vin C

6V

R2

(A) (B) (C) (D)

_ + 2kΩ

P 10kΩ

~

Current - Current feedback Voltage - Voltage feedback Current - Voltage feedback Voltage - Current feedback

10kΩ

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GATE QUESTION BANK

Analog Circuits

EE - 2014 3. In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when μF

If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is_________

( )

IN - 2007 1. A FET source follower is shown in the figure below:

√

+ 15 V

( ) ( )

μF

√

( ) 4.

1

An oscillator circuit using ideal op-amp and diodes is shown in the figure.

The nature of feedback in this circuit is (A) Positive current (B) Negative current (C) Positive voltage (D) Negative voltage ECE/IN - 2013 2. In the feedback network shown below , if the feedback factor k is increased , then the The time duration for +ve part of the cycle is and for ve part is . The value of 5.

Vin

V1

A0

vf=kvout

k

Vout

will be _______________

A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure.

(A) Input impedance increases and output impedance decreases th

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GATE QUESTION BANK

Analog Circuits

(B) Input impedance increases and output impedance also increases (C) Input impedance decreases and output impedance also decreases (D) Input impedance decreases and output impedance increases


6.

[Ans. B] Current Series O/P is taken at collector which is voltage emitter node will be current and resistor is grounded Series

7.

[Ans. A]

[Ans. D] Ideal transconductance amplifier has infinite input and output resistance.

2.

[Ans. A]

3.

[Ans. D]

EE 1. 2

[Ans. A] Output will be either at . When output will be at diode connected to resistance will be on making voltage at point P equal to 6V. When output is at diode connected to 2 resistance will be on making voltage at point equal to .

2

( Refer Diagram below)

5v V +

-

+ -

[Ans. B] It is voltage – voltage feedback.

3.

[Ans. C] Barkhausen criteria for oscillation (for positive feedback)

+

+

4.

2.

[Ans. A] In the parallel RLC Ckt μHand F √

√

A = open loop gain B = feedback factor =

So that for a tuned amplifier, gain is maximum at resonant frequency 5.

(

)

(

)

||

[Ans. B]

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GATE QUESTION BANK

Analog Circuits

2

(

)(

)

(

2

Real and imaginary part on LHS and RHS are same so imaginary part = 0

√

For (+Ve) cycle capacitor charges from 2 2 Apply KCL at node ( ) 4.

[Ans. *] Range 1.2 to 1.3

(

(

)

) 2

( 2

~

) )

( (

)

(

)

(

)

2

2 2 2

2 [

]

For (- Ve) cycle capacitor discharges from 2 2 Apply KCL at node ( ) (

( K th

th

)

)

2 ( 2 ) th


GATE QUESTION BANK

( 2 ) (

2

(

IN 1.

)

[Ans. D] Feedback in this circuit is drain voltage as negative.

) 2 (

2

2.

)

2

[Ans. A] Given connection is voltage series F/B. increases and decreases

2 ] 2

[

Analog Circuits

y q 2 2 [ 5.

(

)

2 ]

[Ans. *] Range 0.62 to 0.66

During the charging of capacitor () ( ) ( ( ) ( )) (

)

() During discharging of capacitor ()

( )

()

(

(

( ( )

( ))

)

) (

)

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GATE QUESTION BANK

Analog Circuits

Operational Amplifiers and Its Applications ECE - 2006 1. For the circuit shown in the following figure, the capacitor C is initially uncharged. At t , the switch S is closed. The voltage across the capacitor at t millisecond is

ECE - 2007 4. For the Op-amp circuit shown in the figure, is 2 kΩ

1 kΩ 1V 1 kΩ

1 kΩ

k

In the figure shown above, the OP AMP is supplied with ±15V. (A) 0 Volts (C) 9.45 Volts (B) 6.3 Volts (D) 10 Volts

(A) (B) 5.

KΩ

k

(A) (B) (C) (D)

k

k

2.

3.

(C) 0.5 V (D) 0.5 V

In the Op-amp circuit shown, assume that the diode current follows the equation I = exp (V/ ). For = 2 V, = and for = 4 V, = . The relationship between and is

Statement for Linked Answer Questions 2 and 3 A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output . Use the component values shown in the figure. k

2V 1V

=√ =e = ln 2 = ln 2

Statement for Linked Answer Questions 6 and 7 Consider the Op-Amp circuit shown in the figure.

In the figure above, the ground has been shown by the symbol The Power dissipation across the transistor Q1, shown in the figure is (A) 4.8 Watts (C) 5.4 Watts (B) 5.0 Watts (D) 6.0 Watts If the unregulated voltage increases by 20%, then power dissipation across the transistor Q1 (A) Increases by 20% (B) Increases by 50% (C) Remains unchanged (D) Decreases by 20% th

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GATE QUESTION BANK

6.

7.

The transfer function

(s)/

(A)

(C)

(B)

(D)

(s) is

10.

Analog Circuits

Consider the Schmidt trigger circuit shown below k

If

sin t and sin t , then the minimum and maximum values of (in radians) are respectively (A) –π/ and π/ (C) –π and (B) and π/ (D) –π/ and

k k

A triangular wave which goes from 12V to 12V is applied to the inverting input of the OPAMP. Assume that the output of the OPAMP swings from +15V to 15V.The voltage at the non-inverting input switches between (A) 12V and +12 V (B) 7.5V and 7.5V (C) 5V and +5V (D) 0V and 5V

ECE - 2008 8. Consider the following circuit using an ideal OPAMP. The I-V characteristics of the diode is described by the relation I= (e

) where m and V is the voltage across the diode (taken as positive for forward bias) D

4k

= 1 100k

ECE - 2009 11. In the following astable multivibrator circuit, which properties of t depend on ?

For an input voltage the output voltage is (A) 0V (C) 0.7V (B) 0.1V (D) 1.1V 9.

The OPAMP circuit shown below represents a

(A) (B) (C) (D)

(A) (B) (C) (D)

High pass filter Low pass filter Band pass filter Band reject filter

th

Only the frequency Only the amplitude Both the amplitude and frequency Neither the amplitude nor the frequency

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GATE QUESTION BANK

Analog Circuits

ECE - 2010 12. Assuming the OP AMP to be ideal, the voltage gain of the amplifier shown below is

13.

(A)

(C)

(

)

(B)

(D)

(

)

The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP AMP and practical diodes)

ECE - 2011 14. The circuit below implements a filter between the input current i and the output voltage v . Assume that the opamp is ideal. The filter implemented is a

i

(A) (B) (C) (D)

low pass filter band pass filter band stop filter high pass filter

ECE/EE/IN - 2013 15. In the circuit shown below what is the output voltage (Vout) if a silicon transistor Q and an ideal op – amp are used?

th

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GATE QUESTION BANK +15 V

Analog Circuits

Q

1 k k

k Vout .7

5V

(A) (B) 16.

15 V

15 V 0.7 V

(C) +0.7 V (D) +15 V

19.

In the circuit shown below the op – amps are ideal. The Vout in Volts is 1 k

1 k +15 V

+15 V

Vout

1 k -15 V

+1V

-15 V

(A) (B)

1 k

1 k

20. (A) 4 (B) 6

(C) 8 (D) 10

ECE - 2014 17. In the low-pass filter shown in the figure, for a cut-off frequency of 5 kHz , the value of (in k ) is ________

k

18.

In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero input offset voltage. The output voltage is

(C) (D)

In the differential amplifier shown in the figure, the magnitudes of the commonmode and differential-mode gains are and , respectively. If the resistance is increased, then

n

In the voltage regulator circuit shown in the figure, the op-amp is ideal. The BJT has .7 and β and the zener voltage is 4.7 V. For a regulated output of 9V, the value of R (in Ω is _________

(A) increases (B) common-mode rejection ratio increases (C) increases (D) common-mode rejection ratio decreases

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GATE QUESTION BANK

21.

Assuming that the Op-amp in the circuit shown is ideal, is given by

Analog Circuits

(A)

(B)

22.

(A)

(C)

(B)

(D)

The circuit shown represents (C)

12V 0.7V

(A) a bandpass filter (B) a voltage controlled oscillator (C) an amplitude modulator (D) a monostable multivibrator

(D)

EE -2006 1. For a given sinusoidal input voltage, the voltage waveform at point P of the clamper circuit shown in figure will be

0.7V 12V

EE -2007 2. The circuit shown in the figure is

~ oad r

Vin

(A) A voltage source with voltage (B) A voltage source with voltage (C) A

current

source

with

current

.

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GATE QUESTION BANK

(D) A

current

source

with

current

Analog Circuits

(D)

. 3.

IC 555 in the adjacent figure is configured as an astable multivibrator. It is enabled to oscillate at t =0 by applying a high input to pin 4. The pin description is: 1 and 8 – supply; 2- trigger; 4- reset; 6 – threshold; 7 – discharge. The waveform appearing across the capacitor starting from t = 0, as observed on a storage CRO is

4.

The switch S in the circuit of the figure is initially closed. It is opened at time t=0. You may neglect the Zener diode forward voltage drops. What is the behaviour of for t ? +10V

+

10K

+10 V

k

8 7

+ IC 555

10K

S

10V

0.01

2, 6 C

4

5.0V

k

3

5.0V

k 1 10V

(A) It

(A)

makes a to at t makes a to at t makes a to at t makes a to at t

(B) It (C) It (D) It (B)

transition . s transition . 7 s transition . s transition . 7 s

from from from from

EE -2008 5. The block diagrams of two types of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block. P

(C)

Q

V0

V0

0

Vin

V0

Vin

V0

Vin 0

Vin

It is desired to make full wave rectifier using above two half – wave rectifiers. The resultant circuit will be

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GATE QUESTION BANK

Analog Circuits

(A) 5V 2.5V t(sec) (B) t(sec)

0 2.5V 5V (C) 5V 0

t(sec)

5V (D) 5V 0 5V

t(sec)

Statement for Linked Answer Questions 7 and 8 A general filter circuit is shown in the figure: 6.

A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangle wave at point ‘ ’ with a peak to peak voltage of 5V for = 0 V.

If the voltage is made + 2.5 V, the voltage waveform at point ‘ ’ will become

7.

If R1 = R2 = RA and R3 = R4 = RB, the circuit acts as a (A) all pass filter (C) high pass filter (B) band pass filter (D) low pass filter

8.

The output of the filter in Q.7 is given to the circuit shown in below figure

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GATE QUESTION BANK

Analog Circuits

/ k z

i

rms

pamp

The gain vs frequency characteristic of output (Vo) will be (A)

k

Gain

0

~

(A) πm (B) πm (C) 1 π m (D) πm

(𝛚)

leading by 0 leading by 0 leading by 900 lagging by 0

(B) 10.

Gain

0

An ideal op-amp circuit and its input waveform are shown in the figures. The output waveform of this circuit will be

(𝛚)

(C)

Gain 6V k

0

(𝛚) +

(D)

3V

k

k

Gain

0

(𝛚) t

EE -2009 9. The following circuit has k C = 10 F . The input voltage is a sinusoid at 50Hz with an rms value of 10V. Under ideal conditions, the current is from the source is th

th

t

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GATE QUESTION BANK

Analog Circuits

(A)

t

t

(B) t t

t

t

t

t

t t

(C) EE -2010 11. Given that the op-amp is ideal, the output voltage is

(D)

(A) 4V (B) 6V

(C) 7.5V (D) 12.12V

EE -2011 12. For the circuit shown below,

13.

the CORRECT transfer characteristic is

th

A-Low pass filter with a cut-off frequency of 30Hz is cascaded with a high pass filter with a cut-off frequency of 20Hz. The resultant system of filter will function as , (A) An all-pass filter (B) An all-Stop filter (C) A band stop (band -reject) filter (D) An band-pass filter

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GATE QUESTION BANK

Analog Circuits

ECE/EE/IN -2012 14. The circuit shown is a (|

|)

utput

ntput

𝛚

(A) Low pass filter with f

rad/s

(B) High pass filter with f

rad/s

(C) Low pass filter with f

rad/s

(D) High pass filter with f

rad/s

𝛚

/ /

EE -2014 15. Given that the op-amps in the figure are ideal, the output voltage is

(|

|)

𝛚

(A) (B) 16.

(C) (D)

/

In the figure shown, assume the op-amp to be ideal. Which of the alternatives gives the correct Bode plots for the transfer function

?

/ /

k

𝛚

/ /

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GATE QUESTION BANK

17. (|

Analog Circuits

An operational-amplifier circuit is shown in the figure.

|)

𝛚

The output of the circuit for a given input is

/ /

( 𝛚

/

)

(

)

/

(

) or

(|

18.

|)

The transfer characteristic of the Op-amp circuit shown in figure is

𝛚

/ /

/

𝛚

/

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GATE QUESTION BANK

2.

Analog Circuits

When the switch S2 is closed the gain of the programmable gain amplifier shown in the following Figure is

k k

(A) 0.5 (B) 2

IN - 2006 1. If the value of the resistance R in the following figure is increased by 50%, then voltage gain of the amplifier shown in the figure will change by

(C) 4 (D) 8

3.

The potential difference between the input terminals of an op-amp may be treated to be nearly zero, if (A) The two supply voltages are balanced (B) The output voltage is not saturated (C) The op amp is used in a circuit having negative feedback (D) There is a dc bias path between each of the input terminals and the circuit ground

4.

A dual op-amp instrumentation amplifier is shown below. The expression for the output of the amplifier is given by

k k Vin

R

(A) 50% (B) 5% (C) 50% (D) negligible amount



(A) v0  1 



th

th

R2   (v  v 1 ) R1  2

th


GATE QUESTION BANK



(B) v0  1 

2R2   (v  v 1 ) R1  2

8.

In the circuit shown in the following figure , the op-amp has input bias current n and input offset voltage m . The maximum dc error in the output voltage is

 2R2 (C) v0  (v  v 1 ) R1 2



(D) v0  1 



Analog Circuits

2R1   (v  v 1 ) R 2  2

k k

5.

An astable multivibrator circuit using a 555 IC is given in the following figure. The frequency of oscillation is 8

I=5 mA

4

+

3

(A) 1.0 mV (B) 2.0 mV

Output 6 2

555 discharge

7

(C= 0.1 uF)

1

(A) 20 kHz (B) 30 kHz

(C) 2.5mV (D) 3.0 mV

(C) 40 kHz (D) 45 kHz

IN - 2007 9. When light falls on the photodiode shown in the following circuit, the reverse saturation current of the photodiode changes from to . k

Statement for linked answer questions 6 and 7 In the Schmitt trigger circuit shown below, the Zener diodes have VZ (reverse saturation voltage) = 6V and VD (forward voltage drop) = 0.7V

.

6.

If the circuit has the input lower trip point (LTP)=0V, then the value of (A) 0.223 (B) 2.67

Assuming the op – amp to be ideal, the output voltage, of the circuit. (A) does not change (B) changes from 1 V to 2V (C) changes from 2 V to 1 V (D) changes from 1 V to 2V 10.

is given as

Consider the linear circuit with and ideal op-amp shown in the figure below. 1

(C) 4.67 (D) ∞

2

’

’

kΩ

7.

The input upper trip point (UTP) of the Schmitt trigger is (A) 1.5 V (C) 2.42V (B) 2.1 V (D) 7V

+

Vi

Vo

The Z-parameters of the two port feedback network are kΩ th

th

th


GATE QUESTION BANK

and = amplifier is (A) + 110 (B) + 11 11.

kΩ. The gain of the

Analog Circuits

(D) o/ i 4 dB

(C) 1 (D) – 120

1dB 20 dB/decade

Consider the circuit shown below

f 330pF 22k Ω

21.9kHz

22k Ω 330pF

12.

10k Ω

In the circuit shown below the switch (S) is closed whenever the input voltage is positive and open otherwise.

7kΩ

R

R

The correct frequency response of the circuit is (A)

. Vin

o/ i 4 dB

The circuit is a (A) Low pass filter (B) Level shifter (C) Modulator (D) Precision rectifier

1 dB 20dB/decade f

21. 9kHz

(B)

13.

/ 0 dB 3 dB

Consider the triangular wave generator shown below.

40dB/decade R kΩ

Input

1 kΩ 10 kΩ

f

21. 9kHz

(C)

o/ i

Assume that the op amps are ideal and have ± 2 V power supply. If the input is a ± 5 V, 50 Hz square wave of duty cycle 50%, the condition that results in a triangular wave of peak to peak amplitude 5 V and frequency 50 Hz at the output is (A) RC = 1 (C)

4 dB 1 dB 40 dB/decade

21. 9kHz

Output

f

(B)

th

(D)

th

th

5


GATE QUESTION BANK

14.

A 555 astable multivibrator circuit is shown in the figure below

Analog Circuits

Noise

Square Wave

7V 8V

KΩ

3V RA

KΩ

RB

output

-7V

threshold

Which one of the following is an appropriate choice for the upper and lower trip points of the Schmitt trigger to recover a square wave of the same frequency from the corrupted input signal (A) ± 8.0 V (C) ± 0.5 V (B) ± 2.0 V (D) 0V

ground

is shorted, the waveform at

is

(A) VC 2/3 VCC cc cc 1/3VCC 0

16.

t

(B)

V ccC 2/3 VCC cc cc 1/3 VCC

cc

The figure shows a signal op-amp differential amplifier circuit . k

k

0

(C)

t

-3V

̅̅̅̅̅̅̅̅̅ tr gger

If

6V

1V -1V

reset discharge

t

cc VC

m

+

+ +

k k

cc

m

cc

2/3 VCC

Which of the following statement about the output is correct? (A) m (B) m m (C) m v m (D) m

1/3 VCC 0

t

(D) VC cc 2/3 VCC 1/3 VCC 0

15.

t

The input signal shown in the figure below is fed to a Schmitt trigger. The signal has a square wave amplitude of 6 V p-p. It is corrupted by an additive high frequency noise of amplitude 8 V p-p.

th

th

th


GATE QUESTION BANK

Common Data Question for Q.No. 17, 18 & 19 Consider the op-amp circuit shown in the figure below:

21.

-+

(A) 10V (B) 10.5V

-

22. -

18.

19.

A differential amplifier shown below has a differential mode gain of 100 and a CMRR of 40dB. If . and . , the output V0 is

k

k

17.

Analog Circuits

If . . and 7 and the op-amp is ideal , the value of the is (A) k (C) k (B) k (D) k Let sin πf t and K . The Op-amp has a slew rate of . / s with its other parameters being ideal .The values of and f for which the amplifier output will have no distortion are, respectively, (A) 0.1 V and 300kHz (B) 0.5 V and 300kHz (C) 0.1 V and 30kHz (D) 0.5 V and 30kHz

(C) 11V (D) 15V

The op-amp circuit shown below is that of a

k

k

k

(A) Low-pass filter with a maximum gain of 1 (B) Low-pass filter with a maximum, gain of 2 (C) High-pass filter with a maximum gain of 1 (D) High-pass filter with a maximum gain of 2 23.

Let and k . Assume that the op-amp is ideal except for a non-Zero input bias current. What is the value of for the output Voltage of the op-amp to be Zero? (A) . k (C) k (B) . k (D) k

In the op-amp circuit shown below the input voltage vin is gradually increased from 10V to +10V. Assuming that the output voltage vout saturates at 10V and +10V,vout will change from

k

IN - 2008 20. An ideal op-amp has the characteristics of an ideal (A) Voltage controlled voltage source (B) Voltage controlled current source (C) Current controlled voltage source (D) Current controlled current source

k

(A) 10V to +10V when vin (B) 10V to +10V when vin (C) +10V to 10V when vin (D) +10V to 10V when vin

th

th

th

1V 1V 1V 1V


GATE QUESTION BANK

24.

For the op-amp circuit shown below approximately equal to

is

27.

Analog Circuits

The input resistance of the circuit shown in the figure, assuming an ideal op-amp, is 2R 3R

R

k

k

(A) (B)

10V 5V

(A) R/3 (B) 2R/3

(C) +5V (D) +10V

IN - 2009 25. The circuit shown is the figure is

28.

(C) R (D) 4R/3

In the circuit shown in the figure, the switch S has been in Position 1 for a long time. It is then moved to Position 2. Assume the Zener diodes to be ideal. The time delay between the switch moving to Position 2 and the transition in the output voltage is osition

k

k

osition

(A) (B) (C) (D) 26.

An all-pass filter A bandpass filter A highpass filter A lowpass filter

In the circuit shown, the Zener diode has ideal characteristics and a breakdown voltage of 3.2V. The output voltage for an input voltage = +1V is closest to

(A) 5.00ms (B) 8.75ms

.7k

ener diode

.7k

ener diode

(C) 10.00ms (D) 13.75ms

IN - 2010 29. In the ideal opamp circuit given in the below figure, the value of Rf is varied from

k

1k to 100k . The gain G = ( ) will

k

k k

k

(A) (B)

10V 6.6V

(C) (D)

(A) (B) (C) (D)

5V 3.2V

th

remain constant at +1 remain constant at -1 vary as -( /10,000 ) vary as ( /10,000) th

th


GATE QUESTION BANK

30.

An active filter is shown in the below figure. The dc gain and the 3dB cut-off frequency of the filter respectively, are, nearly

k

Analog Circuits

IN - 2011 Statement for Linked Answer Questions 33and 34 and in the circuit shown below are matched n-channel enhancement mode MOSFETs operating in saturation mode, forward voltage drop of each diode is .7 reverse leakage current of each diode is negligible and the op-amp is ideal k

R1 = 15.9 k , R2 = 159 k , C1 = 1.0nF (A) 40dB, 3.14 kHz (B) 40dB, 1.00 kHz (C) 20dB, 6.28 kHz (D) 20dB, 1.00 kHz Common Data for Questions: 31 & 32 A differential amplifier is constructed using an ideal op-amp as shown in the adjoining figure. The values of R1 and R2 are 47k and 470k respectively.

31.

32.

The input impedances seen looking into the terminals V1 and V2, with respect to ground, respectively are (A) 47k and 43k (B) 47k and 47k (C) 47k and 517k (D) 517k and 47k

m

33.

The current (A) 1 mA (B) .5 mA

34.

For the computed value of current output voltage is (A) 1.2V (C) 0.2V (B) 0.7V (D) 0.7V

35.

The ideal op-amp based circuit shown below acts as a Ω

in the circuit is (C) m (D) 2 ma

Ω

0.5 µF

, the

0.5 µF

kΩ 1 µF

V1 and V2 are connected to voltage sources having an open circuit output of +1V each and internal resistances of 13k and 3k respectively. The output voltage V0 is (A) 0V (C) 1.5V (B) 0.15V (D) 10V

(A) (B) (C) (D)

th

low-pass filter high-pass filter band-pass filter band-reject filter

th

th


GATE QUESTION BANK

36.

The value of shown below is

of the series regulator

40 V(DC) kΩ

(C) 6.7 V and 4.7 V (D) 5.3 V and 3.3 V 38.

kΩ 6 V(DC)

(A) 24 V (B) 28 V

πt

Assuming base-emitter voltage of 0.7V and β of transistor , the output voltage in the ideal opamp circuit shown below is 5V kΩ

(C) 30 V (D) 32 V

The transfer characteristics of the circuit drawn below is observed on the oscilloscope used in XY mode. The display on the oscilloscope is shown on the right hand side. is connected to the X input with a setting of 0.5 V/div, and is connected to the Y input with a setting of 2 V/div. The beam is positioned at the origin when is zero.

sin

37.

1kΩ

Analog Circuits

Ω

5V

1V

(C) (D)

(A) – 1V (B) 1/3.3V

V V

IN - 2013 39. The operational amplifier shown in the circuit below has a slew rate of 0.8 Volts / s. The input signal is . sin( t . The maximum frequency of input in kHz for which there is no distortion in the output is

~V

470k

22k

0.25sin t

~

V0

(A) 23.84 (B) 25.0 40. Assuming that the op-amp is ideal and the zener diodes have forward biased voltage drop of 0.7V, the values of reverse breakdown voltages of and are, respectively. (A) 3.3 V and 5.3 V (B) 4.7 V and 6.7 V th

(C) 50.0 (D) 46.60

The circuit below incorporates a permanent magnet moving coil milli – ammeter of range 1 mA having a series resistance of 10k. Assuming constant diode forward resistance of 50, a forward diode drop of 0.7 V and infinite reverse diode resistance for each diode , the reading of the meter in mA is

th

th


GATE QUESTION BANK

Analog Circuits R2

mA

10 k

R1

X

10 k

Z Y

~

5V, 50Hz

V0

R3

V0

V1

(A) 0.45 (B) 0.5 41.

R4

(C) 0.7 (D) 0.9 Fig. a +5V

A signal Vi(t)=10+10sin 100πt + 10 sin 4000πt + 10sin 100000πt is supplied to a filter circuit (shown below) made up of ideal op–Amp. The least attenuated frequency component in the output will be

R

R X

. 2k 1k

~

.

Y

.

Strain Gage

750

R

. V1(t)

(A) 0 Hz (B) 50 Hz

Differential Amplifier

R

V0(t)

(C) 2 kHz (D) 50kHz

Statement for Linked Answer Questions 42 and 43 A differential amplifier with signal terminals X, Y, Z is connected as shown in Fig. (a) below for CMRR measurement where the differential amplifier has an additional constant offset voltage in the output. The observations obtained are : when Vi= 2V, V0=3mV, and when Vi=3V, V0=4mV.

42.

43.

th

Fig. b Assuming its differential gain to be 10 and the op – amp to be otherwise ideal , the CMRR is (C) (A) (B) (D)

The differential amplifier is connected as shown in Fig. (b) above to a single strain gage bridge. Let the strain gage resistance vary around its no – load resistance R by ± 1%. Assume the input impedance of the amplifier to be high compared to the equivalent source resistance of the bridge, and the common mode characteristic to be as obtained above. The output voltage in mV varies approximately from (A) +128 to – 128 (C) +122 to – 122 (B) + 128 to – 122 (D) +99 to – 101

th

th


Z

GATE QUESTION BANK

IN - 2014 44. For the op-amp shown in the figure, the bias currents are = 450 nA and =350 nA. The values of the input bias current ( ) and the input offset current ( ) are:

47.

For the given low-pass circuit shown in the figure below, the cutoff frequency in Hz will be ___________. . 7 k

~

k . 7

k

(A) = 800 nA, =50 nA (B) = 800 nA, =100nA (C) = 400 nA, =50nA (D) = 400 nA, 100nA 45.

Analog Circuits

48.

The amplifier in the figure has gain of -10 and input resistance of 50 k . The value of and are

k

The figures show an oscillator circuit having an ideal Schmitt trigger and its input-output characteristics. The time period (in ms) of t is___________.

output

t

k nput

(A) (B) (C) (D) 46.

k

k k

k k

k k

k

Assuming an ideal op-amp in linear range of operation, the magnitude of the transfer impedance

in

of the

current to voltage converter shown in the figure is ___________. k

k . k

i

th

th

th


GATE QUESTION BANK

Analog Circuits


4.

[Ans. C] 2 kΩ

[Ans. D] Initially the switch is closed at t= 0 Then circuit is The circuit is integrator circuit ∫

1 kΩ 1V

dt

1 kΩ

pply K at node dv k dt ( k

2.

)

(

k k k k ( ) k

[Ans. D]

1 kΩ

) . k ) . k

( .

k

k

. k

5.

[Ans. D] Applying KCL

k

exp (

k Volt across 24 k due to virtual ground concept. So voltage across 12k is 3V

)

exp (

k

ln ow

) i

k ln ln

watts 3.

[Ans. B] Unregulated voltage increases by 20% i. e. New regulated voltage = 18V

6.

k k ln

[Ans. A]

watts increases

th

th

th


GATE QUESTION BANK

(

)

(

)

(

)

(

)

10. .

Analog Circuits

[Ans. C] k k

j

k

j j

j

j s s s s 7.

From KCL

s

k

s s

[Ans. C] s s

k

k

.. Threshold depends on output So, when

s s s s

When

s s tan tan tan inimum value of π at maximum value of at

∞

11.

[Ans. A]

12.

[Ans. C] a

8.

[Ans. B] (e

⁄

c

)

Where

Voltage across diode = 60 mV Voltage across 4k resistor s k m Total voltage is m m 9.

b

and k

Op-amp is ideal, so it will satisfy the virtual ground property a c So, we can redraw the circuit as

m .

[Ans. B] At low frequency, capacitor is open and inductor short so, At high frequency capacitor is short and inductor open so, so it is low pass filter.

Circuit is similar to standard inverting amplifier th

th

th


GATE QUESTION BANK

[ 13.

(

Analog Circuits

)]

[Ans. B] i

i

i

i Hence given circuit is a high pass filter

i

15.

[Ans. B]

k .7

Redraw the given circuit Case (1) 5V

i For non-zero value of off, so i

, diode

must be by virtual short equivalent connection for transistor is

Then

t 0 t .7 When , conducting so =0V 14.

both

diode

are 16.

0.7V

[Ans. C] ain of

stage k ( ) k

[Ans. D] At Hence circuit can be redrawn as below

(

1k -2V

k i

m

k ) k

m

i

1k

Vout

i

+1V

-15 V 1k

1k

t ∞ ∞ Hence circuit can be redrawn as below

Gain of stage II =

th

th

*

th

+


GATE QUESTION BANK

17.

[Ans. *] Range 3.1 to 3.26 ut off frequency f

20.

18.

[Ans. B] β e h β is inversily proportional to Re if increases decreases

π

πf . k z

.

Analog Circuits

if

[Ans. *] Range 1092 to 1094

21.

decreases CMRR increases

[Ans. D]

k

k

a .7

Op-Amp ideal so it will satisfy the property of virtual ground .7 ere .

k .7k .7 .7k .7k . 7

(

22.

[Ans. C]

[Ans. D] The circuit shown in the figure has positive feedback. So it can be either a oscillator or multivibrator. So option A and C are cancelled out Now, a voltage controlled oscillator is usually implemented using a artley’s oscillator where the feedback is like shown below.

Due to virtual ground

must have inductors in the feedback circuit. Since the given circuit has no inductors, it has to be a multivibrator.

. 19.

)

.7

o

th

th

th


GATE QUESTION BANK

EE 1.

2.

5. [Ans. D] When is positive, diode will be OFF, so When is negative, diode will be ON, so

Analog Circuits

[Ans. B] From the transfer characteristic of the rectifier P is for for

will be –ve and V will be ve and .7

[Ans. D] Voltage at non inverting terminal, From the transfer characteristics of the rectifier Q is for for

Due to virtual ground current through Load, r r So this circuit acts as a current source with current

3.

[Ans. A] An astable multi-vibrator is providing pulses as given below.

But in this case initial voltage at capacitor is zero so it starts from zero also charging time will be larger (normally) than discharging time but it is made equal by using a diode. 4.

[Ans. D] It is limited circuit It makes transition from +5V to 5V (

e

⁄

For full wave rectifier output is W ‘ ’ must be connected to inverting and ‘ ’ must be connected to non-inverting terminal of the op-amp 6.

[Ans. A] When . , O/P will be clamped by a dc value of 2.5 V

7.

[Ans. C]

)

(Voltage across 100k ) . sec t . 7 s

th

th

th


GATE QUESTION BANK

Analog Circuits

i Let,

only inverting terminal

magnitude i

ere

π

s s is only on non-inverting terminal

Let,

(

π mA and it will be lagging by 90°

) 10. (

(

v

[Ans. D] k

)

)

k

Putting the value of (

, we get k

)

s

ere (

When When

)

upto t after t

ve ve

( ) o it is high pass filter 8.

11.

[Ans. B]

[Ans. D]

( 12. C

)

[Ans. D] First section is differential amplifier having gain off 1.

. . . =

. .

So gain frequency characteristics will be as option (D) 9.

[Ans. D] Voltage at inverting terminal Output is

So th

th

th


GATE QUESTION BANK

(

15.

)

econd stage

Analog Circuits

[Ans. B]

schmitt trigger

x

y

x x y y Applying superposition theorem When y = 0 x

x

when x y/ 13.

[Ans. D] Low Pass Filter

y y

High Pass Filter

x from equation

16.

and

[Ans. A] k i

20Hz

30Hz

Pass band

It is a band pass filter. 14.

[Ans. B] V= =0 = =-

f

= =

= High pass Fitter with f

s rad /sec

th

th

th


GATE QUESTION BANK

17.

Analog Circuits

[Ans. D]

Since positive feedback is there at the output of op-amp 1 depending on output will vary between and At the output of op-amp ⁄ Gain of op-amp is Hence net output will vary between and

IN 1.

[Ans. D] Voltage gain So it does not depend on R.

2.

[Ans. B] When switch (

18.

is closed, )

[Ans. C] 3.

[Ans. C] When Op-Amp is used in ve feedback then voltage difference between the terminals is treated as zero.

4.

[Ans. A] The equivalent circuit is as follows:

-

When is on is off

(

)

(

.

)

when is on is off

------------- (1)

and ⟹

*

+

From equation (1) ⟹

*

+

⟹ =( th

th

) th


GATE QUESTION BANK

5.

[Ans. B] It is a sawtooth wave generator across capacitor C.

8.

Analog Circuits

[Ans. D] DC output Voltage that is output offset voltage, (

)

So maximum offset (dc voltage) at the output will be, (

t T

d o dt d dt d dt

T

T

T

) m m m

m

T 9.

[Ans. B] k

. a

T f

T 6.

k z The given circuit is pply K at ‘a’

[Ans. C] .7 =0 .

When When When reverse circuit of the photodiode changes from 100 to , the output voltage change from 1V to 2V

.

I= I

. =0 .

1.5 =

= 4.47 7.

10.

[Ans. C] 6.7V I 1.5V

I=

.

.

=I =

.

=

. = .

.

[Ans. D] Given that the imp. Parameters are k k k k k k k Then the given op-amp circuit is

.

.

= 2.45V

th

th

th


GATE QUESTION BANK k

k

12.

Analog Circuits

[Ans. D] R

R k

.

k a Vin b

pply K

at ‘a’

pply K

at ‘x’

t

Case-1 When ve half cycle then switch (s) is closed and circuit will become

ain 11.

R

R

[Ans. B] .

p Vin k

k p

( )

k

( 7k

Case -2 When ve half cycle then switch s is open and circuit will become

The given circuit is a second order low pass filter therefore it has a dB/decade and it has 3 dB cut-off frequency as f

)

.

R

R

.

π√ Vin

. k z It has a decade form 0 dB to frequency range.

3 dB in mid

th

th

th


GATE QUESTION BANK

No circuit flow through left side Op-Amp due to open switch so

through resistor . But is shorted out. So discharge time constant .So capacitor will instantly discharge so correct output is in option (A)

= So, the circuit as a rectifier (fullwave) 13.

[Ans. B] O/P of first OPAMP t

15.

t dt, where

∫

k

and O/P of second OPAMP t t ∫

K

t dt

∫

t dt

input is , 50Hz square wave. O/P should be triangular wave of . 50Hz means 20msec, so in 10msec O/P should charge from . to . , when .

Analog Circuits

[Ans. B] The UTP value selected in such a way that to recover a square wave of same frequency from the input signal When The output is When and When We can recover a square wave of same frequency from the corrupted input signal when UTP and LTP value are

16.

[Ans. B] k

t

∫

dt k

m

+

+ +

k

k

m

-

-

14.

[Ans. C]

Apply KCL at KΩ

RA

node

.

reset discharge

KΩ

RB

.

out ̅̅̅̅̅̅̅̅̅ tr gger

.

. . . . . . . Now, Apply KCL at

threshold ground

Capacitor C charges toward through and until in raises upto 2/3 . This voltage is the threshold voltage of pin 6 which drives comparator 1 to trigger the flip-flop so that the output at pin 3 goes low. In addition, the discharge transistor is driven on. Causing the output at pin 7 to discharge the capacitor

.

Node

. For an ideal op-amp . . . . .7 . m

th

th

th

7


GATE QUESTION BANK

17.

[Ans. B]

Analog Circuits

K .

k

k

--

20.

[Ans. A]

21.

[Ans. B]

K K

. K

+

.

and .

-

d

-

Apply the KCL at the inverting node

[ Put,

7

7

.

[

. 7

.

] .

.

and

. 22.

]

.

.

[Ans. D] At low frequency, C is open and At high frequency C is short and (

k 18.

.

.

)

so it is a high pass filter with max. gain of 2.

[Ans. C] sin πf (

)

23.

sin πft d | | dt

πf cos πft

[Ans. D] When and is connected to inverting terminal, output will saturate to +10V.

πf To avoid distortion, Slew rate . f

|

| πf

π f 7 7. 7 So when f k z and Then f 7 7.7 So option (C) is correct. 19.

k

.

k

nd

.

oltage divider rule

When crosses 1 V, So, output will not change to . So, output will change from to when .

[Ans. B] In order to avoid the effect of the bias current, resistance at +ve terminal must be equal to dc resistance seen from –ve terminal by replacing all the sources by their internal resistances. Therefore, k K K th

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GATE QUESTION BANK

24.

[Ans. B] Equivalent circuit k

Analog Circuits

. Now first open the zener diode then circuit will be k

k k k

k

k

By neglecting current through, 100K compare to current through 105 resistance,

. Similarly, by neglecting current through K compare to current through resistance,

So zener will be on and circuit will be as i

k .

i k

.7

k

m

.7 .7

.

o .

will lie between and . Because zener will be on then current 1 mA will be divided in two parts and output will be – 10V < < 3.2V f output then i . m i . m oi i hence this is not possible When . Then i . m . . i . m

.

5.2 V 25.

[Ans. A] s

s

s [

]

So it is all - pass filter. 26.

[Ans. B] Zener diode is in reverse bias. The equivalent circuit is,

here i

k

i

i

k k

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GATE QUESTION BANK

27.

[Ans. A]

Analog Circuits

at time t , when will switch to 5V.

2R

. ,

e ⁄ .7 ms

. t

3R

becomes

R

29.

[Ans. A]

k

Input resistance

a b

⃗

Voltage at inverting terminal, By KCL at node A,

k

and pply K

at a

By KCL at inverting terminal

By equ (1) and (2), Gain = 28.

[Ans. B] When the switch is at position 1, capacitor will be charged by +30V and . so voltage at non –inverting terminal will be 2.5V. When switch is moved to position 2, capacitor will start discharging and when . . will switch to 5V. Equivalent circuit

30.

[Ans. D] DC gain = 20 log | | = 20 dB 3-dB cut-off frequency =

31.

= 1 kHz

[Ans. C] 7 k 7k

k nvertinf terminal

7k 7 k

s

Input impedance seen looking into the terminals and with respect to ground Apply superposition theorem put So voltage at By virtual ground concept voltage at

s

s

c

s voltage i t e

at

i t inverting

e

⁄

terminal

⁄

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GATE QUESTION BANK

So input impedance seen from respect to ground is 7k Now put The input impedance from respect to ground 7 7 k 7k 32.

with

Since, and are all matched transistors So, say Then m or . m lso . m

with

[Ans. B] Now the effective circuit is 7 k

34.

[Ans. A] will be reverse biased while will be forward biased o (Forward voltage drop of ) .7 . .

35.

[Ans. A] The ideal op-amp based circuit is shown in figure 1. with T-network at the input and feedback paths,. Note the specific relations between resistances and capacitances used. It can be shown that this circuit acts as a double integrator and hence it is a low pass filter specifically transfer function

k

7k

Analog Circuits

7 k

Now voltage at point P is 7 7 By virtual ground capacitor, voltage at point Q is 7

k

Now apply KCL at point Q k

.

.

7 k . .

33.

igure

[Ans. B] k

s s

s

s .

or .

m

. sec

s

. s s The frequency response is given by j

which shows the gain falls by

40 dB per decade Clearly

m

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GATE QUESTION BANK

36.

[Ans. C]

When input voltage cycle) everse biased } orward iased .7 .7 .

k

k

38.

Using voltage divider rule

[Ans. D] When input voltage cycle) Forward biased Reverse biased

(During

(During

ve

[Ans. C] .7 . m β

or so

or 37.

Analog Circuits

ve

m

k

.7

ow

39.

[Ans. A] .

f

1kΩ

.

π

π

.

Where f .7 .7 .7 .7

or

. .

40.

[Ans. A]

41.

[Ans. C] For Op-Amp Analysis

.

k z

sin t .

sin t

.7

th

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GATE QUESTION BANK

s s

T.

Analog Circuits

. k z

7 d

. k z

.

d d Lower 3 dB dominant frequency is 0.5 kHz. Total response

ere one pole

s OP-Amp

k

and reject response

is low pass response;

east attenuated frequency k z

Low frequencies z z k z

f . k z

k

.

k z

Only option (C)is lying in this range Option (C) is correct

k z Op-Amp Analysis

42.

[Ans. C] from given data m m Solving (1) & (2)

s s

T.

s s *

+

*

+

43.

[Ans. B] 5V

*

j

+* *

+

j

R

+

R

High pass response R High frequencies

[

. k z

d

7

]

For variation is varies from 12.48mV to 12.56mV

. th

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GATE QUESTION BANK

Analog Circuits

m , given from bridge For change . m =127.8 mV For change op amp . m m . m m . m O/P varies from 7. m to . m

k

i . 47.

[Ans. *]Range 15 to 16 It is second-order low pass system. t’s high cut-off-frequency f

π√ ere

44.

c

[Ans. D] In an OPAMP,

f

45.

n

π√ π

n

48.

[Ans. B] Given k and gain For an inverting amplifier, Gain = -10 / k Then k

f . 7

c

nput bias current n / input offset current n

.

k k

k

. 7 . 7

.

z

[Ans. *] Range 8.0 to 8.5 T

ln [ ln [

/

] ]

. Time period .

46.

.

. m secs

[Ans. 0.6] k

k

i

. k

c

i d

virtual ground K i

at ode

i

k oi i . ki

. k ki ki . k ki . k ki ki . ki ki

k

k

i

ki k

th

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GATE QUESTION BANK

Analog Circuits

Power Amplifiers EE - 2007 1. The input signal Vin shown in the figure is a 1 KHz square wave voltage that alternates between +7V and 7V with a 50% duty cycle. Both transistors have the same current gain, which is large. The circuit delivers power to the load resistor RL. What is the efficiency of this circuit for the given input? Choose the closest answer.

(A) 46% (B) 55%

EE - 2009 2. Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is (A) Pin = Pout for both transformer and emitter follower (B) Pin > Pout for both transformer and emitter follower (C) Pin < Pout for transformer and Pin = Pout for emitter follower (D) Pin = Pout for transformer and Pin < Pout for emitter follower

(C) 63% (D) 92%

Answer Keys and Explanations EE 1.

[Ans. C] load voltage = ±6.3 V So efficiency

2.

[Ans. D] For emitter follower

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GATE QUESTION BANK

Digital Circuits

Number Systems & Code Conversions ECE-2006 1. A new Binary Coded Pentary (BCP) number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 100010011001 corresponds to the following number in base-5 system (A) 423 (C) 2201 (B) 1324 (D) 4231 2.

The number of product terms in the minimized sum-of-product expression obtained through the following K-map is (where, “d” denotes don’t care states) 1 0 0 1 0 d 0 0 0 0 d 1 1 0 0 1 (A) 2 (C) 4 (B) 3 (D) 5

ECE-2007 3. X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is, (A) 100111 (C) 000111 (B) 001000 (D) 101001 ECE-2008 4. The two numbers represented in signed 2’s complement form are P = 11101101 and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2’s complement form is (A) 100000111 (C) 11111001 (B) 00000111 (D) 111111001

ECE-2014 5. The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is __________. EE-2007 1. The octal equivalent of the HEX number AB.CD is (A) 253.314 (C) 526.314 (B) 253.632 (D) 526.632 EE-2014 2. A cascade of three identical modulo-5 counters has an overall modulus of (A) 5 (C) 125 (B) 25 (D) 625 3.

Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter (A) 1 0 0 0 (C) 0 0 1 1 (B) 1 0 0 1 (D) 1 1 0 0

IN-2006 1. A number N is stored in a 4-bit 2’s complement representation as a3 a2 a1 a0 It is copied into a 6-bit register and after a few operations, the final bit pattern is a a a a a 1 The value of this bit pattern in 2’s complement representation is given in terms of the original number is N as (A) 32 a3 + 2N + 1 (C) 2N – 1 (B) 32 a3 – 2N – 1 (D) 2N + 1 IN-2008 2. The result of (45)10 – (45)16 expressed in 6-bit 2’s complement representation is, (A) 011000 (C) 101000 (B) 100111 (D) 101001

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GATE QUESTION BANK

IN-2009 3. The binary representation of the decimal number 1.375 is, (A) 1.111 (C) 1.011 (B) 1.010 (D) 1.001

Digital Circuits

IN-2011 4. The base of the number addition operation 24 + true is (A) 8 (C) (B) 7 (D)

system for the 14 = 41 to be 6 5

Answer Keys & Explanations ECE 1.

2.

3.

4.

5.

[Ans. D] 100 010 4 2

011 3

001 1

[Ans. A] 1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

1 byte = 8 bit so Here d = 7 No. of bits = 28

[Ans. C] x = 01110 y = 11001 ) x y( = 100111 Carry discard it 00111 in 6 bits will be 000111 [Ans. B] igned 2 s complement of P = 11101101 o P = 00010011 igned 2 scomplement of = 11100110 P = P (2 s complement of ) = 00010011 11100110 11111001 ) = 00000111 2 s complement of (P

byte = EE 1.

= 3 5= 4

[Ans. B] Hex number (AB.CD) ⏞ 1010 ⏞ 1011 ⏞ 1100 ⏞ 1101 For finding its octal number, we add one zero in both extreme and group 3 bit together 010 ⏟ 101 ⏟ 011 ⏟ 110 ⏟ 011 ⏟ 010 ⏟ quivalent octal number 253 632

2.

[Ans. C] Overall modulus = 5 = 125

3.

[Ans. D] BCD counter counts up to 1001

IN 1.

[Ans. D] Given number is a a a a in 2’s complement form. We know that in 2’s complement form if we copy MSB at left of MSB any times the number remains unchanged. So a a a a = a a a a a a = When we left shift a number by 1 bit then it is multiplied by 2, a a a a a 0 = 2 Now, a a a a a 0 1=a a a a a 1=2 1

[Ans. *] Range 3.9 to 4.1 A decimal digit is represented by 4 bit in BCD format, so for a decimal number with digits requires 4d bit and th

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GATE QUESTION BANK

2.

[Ans. C] (45) (45)

= ( 24)

3.

[Ans. C] 0.375 × 2 = 0.750 0.750 × 2 = 1.5 1.5 × 2 = 1.0 Hence answer is 1.011

4.

[Ans. B] Let the base is x, Here (24) (14) = (41) (4 x 2 x) (4 x = (4 x 4 + 2x + 4 + x = 4x + 1 x=7

Digital Circuits

= (101000)

1 x) 1 x )

th

th

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GATE QUESTION BANK

Digital Circuits

Boolean Algebra & Karnaugh Maps ECE-2007 1. The Boolean expression ̅̅ ̅ ̅ ̅ ̅̅ be minimized to ̅̅ ̅ (A) ̅̅ ̅ (B)

̅ ̅ ̅

̅ ̅ can

(A) X (B) Y

̅ ̅̅

̅

(C)

ECE-2014 5. The Boolean expression ( ̅̅̅̅̅̅̅̅̅̅̅ ( ̅) ̅ simplifies to

Consider the Boolean function, F(w, x, y, z) = wy + xy + ̅ xyz + ̅ ̅ y + xz + ̅ ̅ ̅. Which one of the following is the complete set of essential prime implicates? (A) ̅̅ (C) ̅ ̅̅ (B) (D) ̅̅

7.

For an n-variable Boolean function, the maximum number of prime implicants is (A) ( ) (C) (B) (D) ( )

ECE-2009 2. If X = 1 in the logic equation

EC/EE/IN -2012 3. In the sum of products function ( ) ∑( ), the prime implicants are ̅ (A) ̅ ̅ ̅ ̅̅ ̅ (B) ̅ (C) ̅ ̅ ̅ ̅̅ ̅ (D) ̅ ̅ ̅

(C) XY (D) X+Y

6.

(D)

[X+Z{ ( +X ) }] { + ( X + Y)} =1 then (A) Y = Z (C) Z = 1 (D) Z = 0 (B) Y =

̅)

)(

EE-2010 Statement for Linked Answer Questions1 and 2 The following Karnaugh map represent a function .

F YZ

ECE-2013 4. In the circuit shown below, Q1 has negligible collector – to – emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is +5 V, X and Y are digital signals with 0 V as logic 0 and as logic 1, then the Boolean expression for Z is

X YZ 0

01

11

10

1

1

1

0

X

00 01 11 1 0 0 1 0 1 1 1 A minimized form of the function is (A) (C)0 1 0 1 (B) (D)

1. R1 Z R2 X

00

F 10 0 0 0

Q1 Diode

2.

Which of the following circuits is a realization of the above function ?

Y

(A) XY

(C) X

(B)

(D) th

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GATE QUESTION BANK

(A)

Digital Circuits

( )

X F (A)

Y Z

X

(B) X

(B) X

F Y Z

(C) X

F Y Z F

(C) X

Y Z

(D) X

4.

The SOP (sum of products) form of a Boolean function is (0,1,3,7,11), where inputs are A,B,C,D (A is MSB, and D is LSB). The equivalent minimized expression of the function is )( ̅ ̅)( ̅ (A) (̅ )( ̅ ) ̅ ̅ ̅ ̅ ̅ )( )( (B) ( )( ) )( ̅ ̅ )( ̅ ̅ ) (C) (̅ )( ̅ ̅)( ̅ ̅)( ̅ (D) (̅ )( )

F Y Z

(D) Y X

( )

F

Y Z

F Z F

EE-2014Y Z 3. Which of the following logic circuits is a realization of the function F whose Karnaugh map is shown in figure

( )

( )

IN-2006 1. Min-term (Sum of Products) expression for a Boolean function is given as follows. f(A, B, C) = m (0, 1, 2, 3, 5, 6) where A is the MSB and C is the LSB. The minimized expression for the function is (A) A + (B  C) (C) (B  C) (B) (A  B) + C (D) IN-2007 2. A logic circuit implements the Boolean function F = ̅ . Y + X .̅ . ̅ . It is found that the input combination X = Y = 1 can never occur. Taking this into account, a simplified expression for F, is given by (C) X + Y (A) ̅ + ̅. ̅ (B) X + Z (D) Y + X. ̅

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GATE QUESTION BANK

3.

Let X and Y = be unsigned 2-bit numbers. The function F = 1 if X > Y and F = 0 otherwise. The minimized sum of products expression for F is (A) + ̅ .̅ .̅ ̅ + ̅ + ̅ (B) ̅ ̅ ̅ ̅ (C) + + ̅ + ̅ .̅ + ̅ (D)

Digital Circuits

IN-2011 6. For the Boolean expression ̅̅ ̅ ̅̅ ̅ ̅ ̅, the minimized Product of Sum (PoS) expression is ( (A) ̅) ( ̅) ̅ ) (B) ( ) (̅ ̅ (C) ( )( ̅) (D)

̅

IN-2008 4. The minimum sum of products form of the Boolean expression ̅̅̅ ̅̅ ̅ Y = ̅̅ ̅ ̅ ̅ ̅ ̅̅ ̅ (A) Y = P ̅ + ̅ ̅ (B) Y = P ̅ ̅ ̅ (C) Y = P ̅ ̅ ̅ ̅ ̅ (D) Y = ̅ ̅ IN-2009 5. The minimal sum-of-products expression for the logic function f represented by the given Karnaugh map is PQ RS 00 01 11 10 00 0 1 0 0

01 11 10

0

1

1

1

1

1

1

0

0

0

1

0

(A) QS + (B) (C)

+ +

(D)

th

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GATE QUESTION BANK

Digital Circuits


4.

[Ans. B] X Y Z 0 0 0

[Ans. D] K-map corresponding to given Boolean expression CD AB

01

00

00

11

10

1 1

01 11

1

0

1

1

1

0

0

1

1

0

Comments Transistor off diode ON Transistor off diode rev biases Transistor ON diode rev biases Transistor ON diode rev biased

1

10

So, ̅Y=Z ̅̅

̅

̅̅

̅̅ ̅

OR ̅

̅

[

]

̅

5. ̅̅

[Ans. A] ( )(

̅̅

( ̅̅

̅̅̅̅ ̅ 6.

2.

[Ans. D] yz

[Ans. D] (

*

wx

̅))+ [

(

(

̅̅

)]

1

1

By putting X = 1 (

* [

̅) (̅̅̅̅̅̅̅̅̅̅̅ ̅) ̅ ̅ ̅) ̅̅̅̅ ̅

))+ [

(

̅(

)]

]

1

1

1

1

1

1

1

1

1

1

xz y

So, P.I. are y, zx, ̅ ̅ 3.

[Ans. A] (

)

̅̅ ̅ ̅

∑(

YZ 1

)

7.

[Ans. D] Maximum P.I will occurs at condition like

̅ 1

1

1 1 ̅ ̅ ( ) So prime implicants are ̅ and ̅ .

1 1

1 1

1 1

1

i.e., no grouping at all so ) So, (

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GATE QUESTION BANK

EE 1.

2.

[Ans. D] ̅

[Ans. B] X

01 11 10

0

1

1

1

0

1

0

0

1

0

F = ̅ ̅ + YZ 2.

3.

4.

[Ans. D] From the figure it is clear that, two NAND ̅ and now two gates generate the ̅ ̅ and inputs AND gates with inputs ̅ Y and Z is used to generate two terms of SOP form and now OR gate is used to sum them and generate the F. [Ans. C] ̅̅ ̅̅ [Ans. A] ̅ ̅ (̅

K – map YZ 00 X 0 0

1

[by consensus theorem]

̅ )( ̅

̅)(

̅)(

x

x

0

X F=Y+X [Ans. D] F = 1 if X > Y, so following will be K – map of function F.

00

00 0

01 0

11 0

10 0

01

1

0

0

0

11

1

1

0

1

10

1

1

0

1

̅ +

F= 4.

A

00

01

11

10

0

1

1

1

0

1 0

1

0

1

F 0 0 1 1 1 0 x x

10 1

1

̅)

[Ans. C] From K – map

BC

Z 0 1 0 1 0 1 0 1

01 11 0 1

̅

3.

IN 1.

̅̅

Truth table: X Y 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1

YZ 00

Digital Circuits

̅ ̅̅̅ +

̅̅̅

[Ans. A]

By K – map

f = ̅ + B ̅ + ̅C =̅+B C th

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GATE QUESTION BANK PQ RS

5.

00 00 1

01 0

11 0

10 1

01

0

0

0

0

11

0

0

0

0

10

1

1

1

1

[Ans. A] PQ RS 00 Q S 00 0

6.

Digital Circuits

01 1

11 0

10 0

1

1

01

0

1

11

1

1

1

0

10

0

0

1

0

[Ans. A] ̅̅ ̅ ̅̅ ̅ ̅ ̅̅ ̅ ̅( ̅ ) ( ̅ ̅̅ ̅ ̅ ̅] ̅̅ [ ̅ )( ̅̅ [( ̅)] ̅̅ ( ̅) ̅̅ ̅ ̅( ̅) ̅ (̅ )(̅ ) ( ̅)( ̅) Alternative method:

0

00 1

01 0

11 0

10 1

0

1

0

1

1

(

̅)(

̅ ̅)

̅)

th

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GATE QUESTION BANK

Digital Circuits

Logic Gates ECE-2007 1. The Boolean function Y =AB + CD is to be realized using only 2 – input NAND gates. The minimum number of gates required is (A) 2 (C) 4 (B) 3 (D) 5

5.

Match the logic gates in Column A with their equivalents in Column B. Column B Column A

ECE-2008 2. The logic function implemented by the following circuit at the terminal OUT is

Q

(A) P NOR Q (B) P NAND Q 3.

X

2

R

3

S

4


P Q Y R

Z

(A) Two or more of the inputs P, Q, R are “0” (B) Two or more of the inputs P, Q, R are “1” (C) Any odd number of the inputs P, Q, R is “0” (D) Any odd number of the inputs P, Q, R is “1”

Y R (A) (B) (C) (D)

Q

ECE-2011 6. The output Y in the circuit below is always “1” when

(C) P OR Q (D) P AND Q

Which of the following Boolean Expressions correctly represents the relation between P, Q, R and ? P Q

1

(A) (B) (C) (D)

OUT P

P

= (P OR Q)XOR R = (P AND Q) XOR R = (P NOR Q) XOR R = (P XOR Q) XOR R

ECE-2010 4. For the output F to be 1 in the logic circuit shown, the input combination should be A

B F C

(A) (B) (C) (D)

A=1, B=1, C=0 A=1, B=0, C=0 A=0, B=1, C=0 A=0, B=0, C=1 th

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GATE QUESTION BANK

Digital Circuits

ECE-2012 7. In the circuit shown 5 Volts A C

(A) (B) (C) (D)

B Y C

10. A

B

̅̅ (

(A) (B)

̅ )

(̅

(C) (D)

4 1024 3 1024 E 2 1024 E

B. C. D.

In the circuit shown in the figure, if the expression for is

̅

(A) (B)

̅

(C) (D)

input S1 S0

0C00H 2C00H 1800H 3800H 0500H 3500H 0800H 2800H

0FFFH, 1C00H 2FFFH, 3C00H 1FFFH, 2800H 3FFFH, 4800H 08FFH, 1500H 38FFH, 5500H 0BFFH, 1800H 2BFFH, 3800H

̅

XOR

1 11 10 01 00

0,

̅

EE-2007 1. A, B, C and D are input bits, and Y is the output bit in the XOR gate circuit of the figure below. Which of the following statements about the sum S of A, B, C, D and Y is correct? A XOR B

1024

A15

A.

8 bit data bus

E

A10 A11 A12 A13 A14

y̅ y̅̅ y y

̅) ̅

ECE-2013 8. There are four chips each of 1024 bytes connected to a 16 bit address bus as shown in the figure below. RAMs 1, 2, 3 and 4 respectively are mapped to addresses

A0 A9

̅y ̅y̅ ̅y̅ ̅y̅̅

Y

E

1FFFH, 3FFFH 2FFFH, 4FFFH 18FFH, 58FFH 1BFFH, 3BFFH

ECE-2014 9. The output F in the digital logic circuit shown in the figure is

C XOR D (A) S is always either zero or odd (B) S is always either zero or even (C) S = 1 only if the sum of A, B, C and D is even (D) S = 1 only if the sum of A, B, C and D is odd

EE-2009 2. The complete set of only those Logic Gates designated as Universal Gates is (A) NOT, OR and AND Gates (B) XNOR, NOR and NAND Gate (C) NOR and NAND Gates (D) XOR, NOR and NAND Gates th

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GATE QUESTION BANK

EE-2011 3. The output Y of the logic circuit given below is

(A) 1 (B) 0

Digital Circuits

average value of the output voltage as function of τ for 0

(A)

t

( )?

Vav 5V

(C) X (D) ̅

τ

T/2

IN-2006 1. All the logic gates in the circuit shown below have finite propagation delay. The circuit can be used as a clock generator, if

(B)

Vav 5V Vav

Y (C) Vav

X (A) X = 0 (B) X = 1

1 T/2

Vav

(C) X = 0 or 1 (D) X = Y

IN-2007 2. Two square waves of equal period T, but with a time delay τ are applied to a digital circuit whose truth table is shown in the following figure. X Y Output 0 0 1 0 1 0 1 0 0 1 1 1 X

5V

T/2 (D)

τ

Vav 2.5V T/2

τ

IN-2009 3. The diodes in the circuit shown are ideal. A voltage of 0V represents logic 0 and +5V represents logic1.The function Z realized by the circuit for inputs X and Y is + 5V

t

T

τ

T/2

Y 1

X

τ

τ T/2

Z

Y

t

The high and the low levels of the output of the digital circuit are 5 V and 0 V, respectively. Which one of the following figures shows the correct variation of the

(C) Z =̅̅̅̅̅̅̅̅ (D) Z=̅̅̅̅

(A) Z=X + Y (B) Z=XY

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GATE QUESTION BANK

IN-2010 4. The logic gate circuit shown in the figure realizes the function A X

Z

Y

(A) XOR (B) XNOR

(C) Half adder (D) Full adder

Digital Circuits

EC/EE/IN-2013 5. A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switch irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) An AND gate (C) A XOR gate (B) An OR gate (D) A NAND gate


5.

[Ans. D] P = ̅̅̅̅̅̅̅ = ̅ ̅ = 4 Q = ̅̅̅̅ = ̅ + ̅ = 2 R = A ⊕ B = A̅ + ̅B = 3 S = A B = AB + ̅ ̅ = 1

6.

[Ans. B] The output Y expression in the ckt (Majority circuit) So that two or more inputs are ‘1’, always ‘1’.

[Ans. B] A B AB + CD C D

2.

3.

[Ans. D] When P = Q = 1, then OUT = 1 P = Q = 0, then OUT = 0 P = 0, Q = 1, then OUT = 0 P = 1, Q = 0, then OUT = 0 So, it is AND gate

7.

[Ans. A] In NMOS circuit Since & are in parallel so those represent ( ) & is in sense, so it represents ‘dot’ operation and the whole function should be inverted or it is complementary logic. So, ̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅ ̅ ̅ ̅ ̅ ( )

8.

[Ans. D] For RAM #1

[Ans. D] X=

;Y=P+Q

Z = XY = =(

. (P + Q) )(

)

⊕ ∴ 4.

⊕

⊕

[Ans. D] For 3 input XNOR for output to be one, two input must be one, and we know that 2- input XOR & XNOR gate are complementary & hence only 1(1’s) will be generated & C=1 is required i.e, When A = 0, B = 0 and C = 1, then F = 1

is

⏟0

0

0

0

⏟1

0 0 0 ⏟ 0

0

⏟ 0 0 0

ower add

⏟ ighest add 0000 ⏟ 1 0 1 1 ⏟ 1⏟1 0 So range of add for RAM #1 0 00 0 which is present only in option D th

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GATE QUESTION BANK

9.

10.

[Ans. A] ( ⊕ )(( ⊕ ) ( ⊕ )(( ⊕ ) ( ⊕ ) ̅) (̅ ̅ ̅

) (

IN 1.

A B

0

[Ans. B] When X = 1, equivalent circuit is

)̅)

Y This circuits act as clock generator.

[Ans. A] C

2. ̅̅̅̅̅̅̅̅

[Ans. C] When τ = 0 X and Y will be same and out-put will be equal to dc of 5V.

1 ̅̅̅̅̅̅̅̅

0

( ( ̅

EE 1.

2.

3.

of each other and output will be equal to dc 0.

AB

̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅

y

When τ = , X and Y will be complement

̅̅̅̅̅̅̅̅

A B

) ̅̅̅̅ ) (̅ ̅

̅)

[Ans. B] ⊕ ⊕ ⊕ from the given diagram. We know that sum of any number of bits is XOR of all bits. So S ⊕ ⊕ ⊕ ⊕ S=Y⊕Y S = either zero or even because LSB is zero (always). [Ans. C] NOR and NAND are designated as universal logic gates, because using any one of them we can implement all the logic gates. [Ans. A] Y= .̅ X 1 0

0 1

̅. ̅ Y

̅

Digital Circuits

When τ increases from 0 to

, O/P will

decrease from 5V to 0V linearly. 3.

[Ans. B] When any of X or Y is zero, Z = 0. For X = Y = 1, Z = 1

4.

[Ans. A] X

y x

Y

y

Z= y. y 5.

Z

= y

y

y =x⊕y

[Ans. C] When both switches in on position, bulb is off When both switches in off position, bulb is off Bulb 0 0 0 0 1 1 1 0 1 1 1 0 It is a XOR gate

1

1 1

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GATE QUESTION BANK

Digital Circuits

Logic Gate Families ECE - 2007 1. The circuit diagram of a standard TTL NOT gate is shown in the figure. When Vi = 2.5V, the modes of operation of the transistors will be

1 4kΩ 4kΩ

Q1

Ignoring the body-effect, the output voltages at P, Q and R are,

100kΩ

R 2

(A) 4 V, 3 V, 2 V (B) 5 V, 5 V, 5 V

Q4 D

4.

Q2

(C) 4 V, 4 V, 4 V (D) 5 V, 4 V, 3 V

The output (Y) of the circuit shown in the figure is

Q3 1kΩ -

̅

(A) Q1: reverse active; Q2: normal active; Q3: saturation; Q4: cut-off (B) Q1: reverse active; Q2: saturation; Q3: saturation; Q4: cut-off (C) Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation (D) Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active ECE - 2009 2. The full forms of the abbreviations TTL and CMOS in reference of logic families are (A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor (D) Tristate Transistor Logic and Complementary Metal Oxide Silicon

̅

(A) ̅ (B)

̅ ̅ ̅

̅

(C) ̅ (D)

̅ ̅

EE - 2010 1. The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10ns. The output Y of the circuit is x 100 ns

1

ECE - 2014 3. In the following circuit employing pass transistor logic, all NMOS transistors are identical with a threshold voltage of 1V.

0

th

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GATE QUESTION BANK X

Y

Digital Circuits

IN - 2007 1. A CMOS implementation of a logic gate is shown in the following figure: 5v

(A) Y

X Y

PMOS

1 0

t

NMOS

(B) Y

The Boolean logic function realized by the circuit is. (A) AND (C) NOR (B) NAND (D) OR

1 0

t

(C)

IN - 2014 2. The figure is a logic circuit with inputs A and B and output Y. = + 5 V. The circuit is of type

Y

1

t

0

(D)

Y

1 0

t

(A) NOR (B) AND

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(C) OR (D) NAND

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GATE QUESTION BANK

Digital Circuits


EE 1.

[Ans. B] reverse active,

saturation

saturation, 2.

[Ans. A] X

Y

cut off

[Ans. C] TTL – Transistor Transistor Logic CMOS – Complementary Metal Oxide Semiconductor

A

B

1 X 0

3.

[Ans. C] Suppose all NMOS at saturation

A

B

For 1 &

Y=X

B

4 For (

IN 1.

1)

(4

[Ans. C] NOR Gate

)

4

(4

)

2.

[Ans. D] Given circuit is of the standard 2 input NAND gate.

4 1 (4

4.

)

4 4

[Ans. A] The given circuit is CMOS implementation If the NMOS is connected in series, then the output expression is product of each input with complement to the final product. ̅̅̅̅̅̅̅̅̅ ̅ ̅

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GATE QUESTION BANK

Digital Circuits

Combinational and Sequential Digital Circuits ECE - 2006 1. The point P in the following figure is stuck-at-1. The output f will be A B

1 0 1 1 MSB

A

Q

D

LSB

FULL ADDER

CK f

0 0 1 1 Shift Registers

P

B

Q

D

S

0

Ci

0

Q

C0

CK D

C

(A) ̅̅̅̅̅̅̅ (B) ̅

0

CK

̅ (C) (D) A CLOCK

2.

Two D-flip-flops, as shown below, are to be connected as a synchronous counter that goes through the following Q1Q0 sequence ………… The inputs and respectively should be connected as

̅̅̅̅

(A) S =0, C0= 0 (B) S = 0, C0= 1

ECE - 2007 4. For the circuit shown, the counter state (Q1 Q0) follows the sequence

̅̅̅̅

c

c

(A) (B) (C) (D) 3.

(C) S = 1, C0= 0 (D) S = 1, C0= 1

̅̅̅̅ and ̅̅̅̅ and ̅̅̅̅ and ̅̅̅̅ ̅̅̅̅ ̅̅̅̅and

For the circuit shown in figure below, two 4-bit parallel-in serial-out shift registers loaded with the data shown are used to feed the data to a full adder. Initially, all the flip-flops are in clear state. After applying two clock pulses, the outputs of the full adder should be

(A) (B) (C) (D) 5.

……… ……… ……… ………

The following binary values were applied to the X and Y inputs of the NAND latch shown in the figure in the sequence indicated below: X=0, Y=1; X=0, Y=0; X=1, Y=1 The corresponding stable P, Q outputs will be X P

Q

Y th

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GATE QUESTION BANK

(A) P=1, Q=0; P=1, Q=0; P=1, Q=0 or P=0, Q=1 (B) P=1, Q=0; P=0, Q=1 or P=0, Q=1; P=0, Q=1 (C) P=1, Q=0; P=1, Q=1; P=1, Q=0 or P=0, Q=1 (D) P=1, Q=0; P=1, Q=1; P=1, Q=1

Digital Circuits

1

1

1

1

CLK

1 CLK 0 T

6.

In the following circuit, X is given by 0 1 1 0

I0 I1 I2 I3

4-to-1 MUX Y S1

S0

A

B

0 1 1 0

I0 I1 I2 I3 S1

4-to-1 MUX Y

t

t

Which of the following waveforms correctly represents the output at ?

X

S0

(A)

C

1 0 2T

t 1

(A)

(B)

(B) (C)

0 4T t

(D)

(C)

ECE - 2008 7. For the circuit shown in the following figure, are inputs to the 4:1 multiplexer. R(MSB) and S are control bits.

1 0 2T

t

(D)

1 0 4T t

z

9.

For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible

̅

̅

The output Z can be represented by (A) PQ + P ̅ S + ̅ ̅ ̅ (B) P ̅ +PQ̅ +̅ ̅ ̅ (C) P ̅ ̅ + ̅QR +PQRS + ̅ ̅ ̅ (D) PQ ̅ +PQR̅ +P ̅ ̅S + ̅ ̅ ̅ 8.

Which of the following statements is true? (A) Q goes to 1 at the CLK transition and stays at 1. (B) Q goes to 0 at the CLK transition and stays at 0. (C) Q goes to 1 at the CLK transition and goes to 0 when D goes to 1. (D) Q goes to 0 at the CLK transition and goes to 1 when D goes to 1.

For each of the positive edge – triggered J-K flip flop used in the following figure, the propagation delay is T.

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GATE QUESTION BANK

ECE - 2009 10. What are the counting states (Q1, Q2) for the counter shown in the figure below?

c

(A) (B) (C) (D) 11.

ip

Statement for Linked Answer Question 13 and 14. Two products are sold from a vending machine, which has two push buttons and . When a button is pressed, the price of the corresponding product is displayed in a 7-segment display. If no buttons are pressed ‘ ’ is displayed, signifying ‘ s. ’ If only is pressed, ‘ ’ is disp ayed signifying ‘ s. ’ If only p are pressed, ‘5’ is disp ayed signifying ‘ s.5’ If both a d are pressed, ‘E’ is displayed, Signifying ‘Err r’. The names of the segments in 7-segment display, and the glow of the display for ‘ ’ ‘ ’ ‘5’ a d ‘E’ are sh w be w.

ip

11, 10, 00, 11, 10 . . . . . 01, 10, 11, 00, 01 . . . . . 00, 11, 01, 10, 00, . . . . . 01, 10, 00, 01, 10 . . . . . .

Refer to the NAND and NOR latches shown in the figure. The inputs for both the latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs ( ) are P1

Q1 P 1

Digital Circuits

a f Q1

0

g

e

2

5

E

b c

d P2

Q2 P 2

(A) NAND : first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0) (B) NAND : first (1, 0) then (1, 0) NOR: first (1, 0) then (1, 0) (C) NAND : first (1, 0) then (1, 0) NOR: first (1, 0) then (0, 0) (D) NAND : first (1, 0) then (1, 1) NOR: first (0, 1) then (0, 1) 12.

What are the minimum number of 2 to. 1 multiplexers required to generate a 2-input AND gate and a 2-input Ex-OR gate? (A) 1 and 2 (C) 1 and 1 (B) 1 and 3 (D) 2 and 2

Q2

Consider (i) Push Button pressed/not Pressed in a equivalent to logic 1/0 respectively. (ii) A segment glowing / not glowing in the display is equivalent to logic 1/0 respectively. 13.

If segment a to g are considered as functions of and , then which are of the following is correct? (A) g ̅ d c e (B) g d c e ̅ (C) g e b c (D) g e b c

14.

What are the minimum numbers of NOT gates and 2-input OR gates required to design the logic of the driver for this 7-segment display? (A) a d (B) a d (C) a d (D) a d

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GATE QUESTION BANK

ECE - 2010 15. The Boolean function realized by the logic circuit shown is

Data

Digital Circuits

D

Q

D

Y

Q

Clock

(A) (B) (C) (D)

MUX

19. (A) (B) (C) (D) 16.

∑ ∑ ∑ ∑

5 5 5

cha ged fr “ ”t “ ” cha ged fr “ ”t “ ” changed in either direction not changed

The logic function implemented by the circuit below is (ground implies a logic “ ” MUX

5

5

Assuming that all flips flops are in reset condition initially, the count sequence observed at in the circuit shown is utput

(A) F = AND(P, Q) (B) F = OR(P, Q)

̅

̅

̅

c

(A) (B)

… …

(C) (D)

… …

ECE - 2014 20. Five JK flip-flops are cascaded to form the circuit shown in Figure. Clock pulses at a frequency of 1 MHz are applied as shown. The frequency (in kHz) of the waveform at Q3 is _____.

ECE - 2011 17. Two D flip – flops are connected as a synchronous counter that goes through the following sequence The connections to the inputs are (A)

and

(C) F = XNOR(P, Q) (D) F = XOR(P, Q)

c

c

c

c

c

c

21.

The digital logic shown in the figure satisfies the given state diagram when Q1 is connected to input A of the XOR gate.

(B) (C) (D) ̅̅̅̅

18.

̅̅̅̅

When the output Y in the circuit below is “ ” it i p ies that data has th

th

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GATE QUESTION BANK

Suppose the XOR gate is replaced by an XNOR gate. Which one of the following options preserves the state diagram? (A) Input A is connected to ̅̅̅̅ (B) Input A is connected to (C) Input A is connected to ̅̅̅̅ complemented (D) Input A is connected to ̅̅̅̅ 22.

23.

Digital Circuits

(A) (B)

(C) (D)

and S is 24.

In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X Y) are given by (A) (B) ̅ (C) ̅̅̅̅̅̅̅̅ ̅ (D) In the circuit shown, choose the correct timing diagram of the output (y) from the given waveforms a d .

The outputs of the two flip-flops Q1, Q2 in the figure shown are initialized to 0, 0. The sequence generated at Q1 upon application of clock signal is

̅̅̅̅

̅̅̅̅

(A) (B) 25.

… …

(C) (D)

… …

The circuit shown in the figure is a E

̅

atch

̅

E

atch

̅

utput y

(A) Toggle Flip Flop (B) JK Flip Flop (C) SR Latch (D) Master-Slave D Flip Flop ̅

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GATE QUESTION BANK

26.

Digital Circuits

Consider the multiplexer based logic circuit shown in the figure.

Which one of the following Boolean functions is realized by the circuit? ̅ ̅̅̅ (A) (B) (C) (D) 27.

In the circuit shown, W and Y are MSBs of the control inputs. The output F is given by

(A) (B) (C) (D)

28.

̅

29.

̅ ̅ ̅̅ ̅ ̅ ̅ ̅̅ ̅ ̅ ̅ ̅ ̅̅

If X and Y are inputs and the Difference (D = X – Y) and the Borrow (B) are the outputs, which one of the following diagrams implements a half-subractor?

An 8-to-1 multiplexer is used to implement a logical function as shown in the figure. The output Y is given by

30.

th

̅ ̅

̅

(A) (B) (C) (D)

̅ ̅

̅

̅̅

̅

A 16-bit ripple carry adder is realized using 16 identical full adders (FA) as shown in the figure. The carrypropagation delay of each FA is 12 ns and the sum-propagation delay of each FA is 15 ns. The worst case delay (in ns) of this 16-bit adder will be __________

th

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GATE QUESTION BANK

Digital Circuits

EE - 2011 3. A two-bit counter circuit is shown below.

EE - 2006 1. A 4 1 MUX is used to implement a 3 – input Boolean function as shown in figure. The Boolean Function F(A,B,C) implemented is

B

(A) (B) (C) (D)

F (A,B,C) = F (A,B,C) = F (A,B,C) = F (A,B,C) =

J

Q

K

̅

K

̅

If the state of the counter at the clock time t is “ ” the the state of the counter at t (after three clock cycles) will be (A) 00 (C) 10 (B) 01 (D) 11

F (A, B, C)

‘ ’

Q

CLK

A

‘ ’

J

EE - 2013 4. The clock frequency applied to the digital circuit show in the figure blow is 1 kHz. If the initial state of the output Q of the flip – f p is ‘ ’ the the freque cy f the output wavefrom Q in kHz is

C

(1,2,4,6) (1,2,6) (2,4,5,6) (1,5,6)

X

EE - 2008 2. A 3 line to 8 line decoder, with active low outputs, is used to implement a 3 – variable Boolean function as shown in the figure.

CLK

3L x 8L Decoder

z y x

0 1 2 3 4 5 6 7

(A) 0.25 (B) 0.5 F

The simplified form of Boolean function p e e ted i ‘ r duct f u ’ f r wi be (A) (X + Z). (̅ ̅ ̅ ). (Y + Z) (B) (̅ ̅ ). (X + Y + Z). (̅ ̅ ). (C) (̅ ̅+ Z). (̅ + Y + Z). (X + ̅ + Z). (X +Y + ̅ ) ̅ ). (X +̅ + ). (D) (̅ ̅ . (̅ ̅ ̅ ( )

T

Q

>

(C) 1 (D) 2

EE - 2014 5. A state diagram of a logic gate which exhibits a delay in the output is shown in the figure, where X is the d ’t care condition, and Q is the output representing the state. ⁄ ⁄

The logic gate represented by the state diagram is (A) XOR (C) AND (B) OR (D) NAND th

th

th


Q

GATE QUESTION BANK

6.

Digital Circuits

A 3-bit gray counter is used to control the output of the multiplexer as shown in the figure. The initial state of the counter is . The output is pulled high. The output of the circuit follows the sequence -bit gray c u ter

f ip f p

5

̅ ̅ E

8.

Two monoshot multivibrators, one positive edge triggered ( ) and another negative edge triggered ( ), are connected as shown in figure

utput

(A) (B) (C) (D) 7.

5

̅

A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.

̅

The monoshots a d when triggered produce pulses of width a d respectively, where . The steady state output voltage of the circuit is

f ip f p t

̅

f ip f p

̅

f ip f p ̅

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GATE QUESTION BANK

IN - 2006 1. Given that the initial state (Q1Q0) is 00, the counting sequence of the counter shown in the following figure is, Q1Q0 =

Digital Circuits

IN - 2007 Statement for Linked Answer Questions 3 & 4 Consider the circuit shown in the following figure.

‘ ’

J0

Q

J1

Q1

̅

K

̅1

0

1

0

CLK

(A) (B) (C) (D) 2.

K0

00 00 00 00

11 01 11 10

01 11 10 01

10 10 01 11

00 00 00 00

A combinational circuit using a 8-to-1 multiplexer is shown in the following figure. The minimized expression for the output (Z) is 1 1 1 0 0 0 1 0

I0 I1 I2 I3 I4 I5 I6 I7

(MSB)

The correct input-output relationship between Y and ( , ) is (A) Y= + (C) Y= ⨁ (B) Y= (D) Y= ⨁

4.

The

flip-flops are initialized =000. After 1 clock cycle, is equal to (A) 011 (C) 100 (B) 010 (D) 101

5.

A sequential circuit is shown in the figure below. Let the state of the circuit be encoded as . he tati implies that state Y is reachable from state X in a finite number of clock transitions.

Y Z

MUX

A B C Select Inputs

̅ ̅ (A) (B) C (A + B)

3.

(LSB)

̅̅ (C) ̅ (D) ̅ + AB

D

̅A

to

QB

QA

̅

Q

̅

CLK

TA

CLK

Q TB

CLK

Identify the INCORRECT statement. (A) (C) (B) (D)

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GATE QUESTION BANK

6.

A MUX circuit shown in the figure below implements a logic function . The correct expression for is Z

1 MUX out

̅

9.

The output F of the multiplexer circuit shown below expressed in terms of the inputs P, Q and R is

F1

R

I0

̅

I1 4

0 S

X

Digital Circuits

̅

I2

R

I3

1 MUX out

̅

F0

0 Y

(A) (B) (C) (D)

S

(A) (̅̅̅̅̅̅̅̅)⨁ ̅̅̅̅̅̅̅̅) ⨁ (B) (̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅

⨁ ⨁̅ ⨁ +Z

(C) (D)

IN - 2008 Statement for Linked Answer Questions 7 &8 Consider the counter circuit shown below.

10.

F

Y

1 MUX

S1

S0

P

Q

F=P⨁Q⨁R F = PQ + QR + RP F = (P ⨁ Q) R F = (P ⨁ Q) ̅

The inverters in the ring oscillator circuit shown below are identical. If the output waveform has a frequency of 10 MHz, the propagation delay of each inverter is Output

l

J Q

Q0

J Q

Q1

J

Q

Q2

Clock KCLR

KCLR

KCLR

̅̅̅̅̅̅̅

J Q Q3

(A) 5 ns (B) 10 ns

K CLR

̅̅̅̅̅̅̅

IN - 2009 11. The figure below shows a 3-bit ripple counter, with as the MSB. The flipflops are rising-edge triggered. The counting direction is

8.

J

1 Clock

Y

7.

(C) 20 ns (D) 50 ns

In the above figure, Y can be expressed as (A) (C) (B) (D)

Q

1

CLK 1

K

Q

J

Q1

1

1

K

Q

CLK

CLK ̅

J

̅

1

K

̅

(A) always down (B) always up (C) up or down depending on the initial state of only (D) up or down depending on the initial states of and

The above circuit is a (A) Mod – 8 Counter (B) Mod 9 Counter (C) Mod – 10 Counter (D) Mod – 11 Counter

th

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Q2

GATE QUESTION BANK

12.

In the figure shown, the initial state of Q is 0. The output is observed after the application of each clock pulse. The output sequence at Q is 1

CLK

In this circuit, the race around (A) Does not occur (B) Occurs when CLK = 0 (C) Occurs when CLK = 1 and A =B =1 (D) Occurs when clk = 1 and A = B = 0

̅

K

(A) 0 0 0 0 . . . (B) 1 0 1 0 . . .

(C) 1 1 1 1 . . . (D) 1 0 0 0 . . .

16.

IN - 2011 13. The circuit below shows as up/down counter working with a decoder and a flip-flop. Preset and clear of the flip-flop are asynchronous active-low inputs

̅

Consider the given circuit

Q

J

CLOCK

15.

Digital Circuits

̅ ̅

̅

̅

̅

̅

The state transition diagram for the logic circuit shown is

x ̅

x

e ect

̅

3 to 8 Decoder C ̅̅̅̅̅̅̅̅ reset Q Flip-Flop Clock ̅̅̅̅̅̅̅ c ear ̅

D

Count Down

B

A(L.SB)

Up/down Counter

Count Up Clock

Assuming that the initial value of counter output ( as zero, the counter output in decimal for 12 clock cycles are (A) 0,1,2,3,4,4,3,2,1,1,2,3,4, (B) 0,1,2,3,4,5,0,1,2,3,4,5,0, (C) 0,1,2,3,4,5,5,4,3,2,1,0,1 (D) 0,1,2,3,4,5,4,3,2,1,0,1,2 ECE/EE/IN - 2012 14. The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is (A) 4 (C) 8 (B) 6 (D) 10 th

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GATE QUESTION BANK

IN - 2013 17. The digital circuit shown below uses two negative edge- triggered D flip- flops. Assuming initial condition of and as zero, the ouput of this circuit is

D1

D0

D-Flip-flop

D-Flip-flop ̅̅̅̅

̅̅̅̅

Digital Circuits

IN - 2014 18. Frequency of an analog periodic signal in the range of 5 kHz - 10 kHz is to be measured with a resolution of 100Hz by measuring its period with a counter. Assuming negligible signal and transition delays the minimum clock frequency and minimum number of bits in the counter needed, respectively, are: (A) 1 MHz, 10-bits (B) 10 MHz, 10-bits (C) 1 MHz, 8-bits (D) 10MHz, 8-bits

Clock

(A) (B) (C) (D)

00, 01, 10, 11, clock 00, 01, 11, 10, 00, 11, 10, 01, 00, 01, 11, 11,

…… …… …… ……


S = 1, [Ans. D] When P is stuck at 1, f = A

4.

[Ans. B] The i/p to first F/F = ̅̅̅̅̅̅̅̅̅̅̅ The i/p to second F/F =

f

B

0

C (

2.

0 0 1 1

3.

)

[Ans. A] Present State

So

,C=1

0 1 1 0

[(

0

0

1 1 0 0

0 1 1 0

1 1 0 0

5.

[Ans. C] When X = 0, Y = 1 then P = 1 and Q = 0 X = 0, Y = 0 then P = 1 and Q = 1 X = 1, Y = 1 then P = 0 and Q = 1 or P = 1 and Q = 0

6.

[Ans. A] Let the output of first MUX is Y Y = ̅B + A̅ = A ⨁ B

and

[Ans. D] Before clock pulse S = C = 0 After first clock pulse A = B = 1 So S = 0, C=1 After second clock pulse A = B = 1 So

(initially at rest)

1st 0 0 1 1 clk 2nd 1 1 0 0 3rd 0 0 0 0 So sequence generated 00, 01, 10, 00

Next State 0 1 1 0

0

)]

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GATE QUESTION BANK

̅ X=̅ ⨁ So X = A ⨁ B ⨁ C ̅ ̅ ̅̅ i.e, ̅ ̅ 7.

So sequence is (00, 11, 10, 00, 11) or …..

[Ans. A] (

ts

11.

[Ans. C] For NAND latch, when ( ) are (0, 1), ( ) will be (1, 0) and when ( ) are (1, 1), ( ) will be (1, 0). For NOR latch, when ( ) are (0, 1), ( ) will be (1, 0) and when ( ) are (1, 1), ( ) will be (0, 0).

12.

[Ans. A]

)

̅̅ ̅̅

̅ ̅̅̅

̅ ̅

̅ ̅

ap PQ

RS

00 00

01

1

11 10 1

1

01

1

1

11

1

1

10

1

Digital Circuits

̅ ̅̅ SP

O Y = AB

PQ

B

̅̅̅

So Now only option (A) has two similar terms, we can say that option A is not most simplified version of Z it can be obtained as

A B Y=A ⨁B 1

PQ RS

11

10

1

1

01

1

1

11

1

1

10

1

00 00

01

1

̅̅ ̅

0 A

P̅S

B

PQ

So option A is correct choice 8.

9.

10.

13.

[Ans. B] At clock will be divide by 4 and will have 2 T delay w.r.t clock. [Ans. C] Initially, when clk is high and D is low, Q = 0 or 1, When clk goes low and D is also low , Q = 1, When clk is low and D goes high, Q = 0.

5

0 1 0 1 1

0 1 0 0 1

0 1 1 1 1

0 1 1 0 1

0 1 0 0 1

14.

a

b

c

d

e

f

g

0

0

1

1

1

1

1

1

0

0

1

1

0

1

1

0

1

1

1

0

1

1

0

1

1

0

1

1

1

1

0

0

1

1

1

1

a b c d e f g g d

[Ans. A]

0 1 0 1 1

[Ans. B]

̅̅̅̅ ̅̅̅̅

…. …. c e ̅̅̅̅ … ̅̅̅̅ … … c

e

[Ans. D] gates gates

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GATE QUESTION BANK

15.

[Ans. D] ̅̅ ̅̅

̅

̅̅

̅

̅̅ ̅

̅

̅̅

̅̅ ̅

21.

[Ans. D] If one input of the gate is kept constant, and A is interchanged with ̅, an XNOR gate acts as XOR gate.

22.

[Ans. C] In half subtractor Difference x y xy x y b rr w x y x y Borrow 0 0 0 0 1 1 1 0 0 1 1 0

̅̅ ̅

̅̅ [ ̅ ̅̅

x

x̅

x.

x ̅ ̅ ̅̅̅

]

̅̅

5 16.

[Ans. D] ⨀

=

Initially 0 After first clk 1 nd After 2 clk 1 After 3rdclk 0 So . . . . . .. 17.

[Ans. D] Q(present) 0 1 0 1

0 1 1 0

Digital Circuits

0 0 1 1

0 0 0 1

23.

Difference 0 1 1 0

[Ans. C]

̅

Q (next) 1 0 1 0

y

1 1 0 0

1 0 1 0

1 1 0 0

̅

⨀ ̅̅̅̅ ̅̅̅̅ ̅ 18.

19.

20.

[Ans. A] Given Y=1, this implies Both the output of D- flip flop should be 1 i.e, input at first flip flop is 1 and for output of 2nd flip flop to be 1, inverted output of first flip should be 1 in previous clock, for which input must be 0 so data is changing from 0 to 1. [Ans. D] From the CKT 0 is connected to & d ‘ ’ is c ected t

So wave form 24.

[Ans. *] Range 62.4 to 62.6 .5

[Ans. D] This is a figure of Johnson counter So 0 1 1 0 0 1 So =

and

is correct

0 0 1 1 0 0

z th

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GATE QUESTION BANK

25.

[Ans. D]

26.

[Ans. D]

2.

Digital Circuits

[Ans. A] ∑

5

[ 3.

(

[Ans. C] Clock

)

]

Input

Output

̅

̅

utput f first et utput f sec d

Initial state 1 2 3

̅ ̅

28.

[Ans. C] ̅ ̅̅ ̅ ̅

X

̅̅̅ ̅̅

̅ ̅

̅̅ CLK

[Ans. A]

30.

EE 1.

[Ans. C] ̅̅ ̅

1 1 0

̅̅ ̅

1 0 1

1 0 0

[Ans. B]

From above Fig. ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅ . ̅̅̅̅̅̅̅̅̅̅̅ ̅ ] [ ̅ ecause ̅̅̅̅̅̅̅̅̅ ̅ a ways

̅ 29.

0 1 0

0

̅ 4.

27.

1 0 1

1

̅

T

Q

Q

>

a ways

̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅ . ̅̅̅̅̅̅̅̅̅̅̅ ̅ ̅̅̅̅ ̅ .

[Ans. *] Range 194.9 to 195.1 Worst case propagation delay = carry propagation delay of 15FA stages + max(carry Pd & sum Pd of last FA stage) 5 5 5 s

i put a ways r ‘ ’ f ip flop if input is =1 then output will be complemented at the time of triggering. CLK

T

[Ans. A]

2T

f

F (A, B, C)= ̅̅ ̅̅

̅̅ ̅̅

̅ ̅

̅ ̅ ̅

th

.5 f .5 .5 z

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GATE QUESTION BANK

5.

[Ans. D]

IN 1.

Digital Circuits

[Ans. A] Clock

x

( 1 1 1 1 1

gate 6.

[Ans. A] Gray code output

o/p 0 0 1 1

0 1 0 1

̅ E

When

0 0 0 0 1 1 1 1

0 0 1 1 1 1 0 0

0 1 1 0 0 1 1 1

2.

1 0 1 0

0 1 0 1 0

0 1 1 0 0

[Ans. C]

(

) )(

)

utput e ect i puts 3.

[Ans. B] Y=

[Ans. A] T-FF to JKFF is given by eq ̅

4.

[Ans. B] a d

This is implemented on options [A] 8.

1 0 1 0

(

utput 7.

1 1 1 1 1

)

[Ans. C] Let is initially high. Since is from will be high for time. The waveform is shown below:

So initially

it means so after one clock cycle will be 010.

5.

[Ans. D] 01 10, both both will toggle.

̅

= 1 so

and

t

6.

[Ans. B]

7.

[Ans. A] ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ Y = ̅̅̅̅̅̅̅ . ̅̅̅̅̅̅̅

trigger t

t

= = t

8.

+

[Ans. C] Whenever Y=1, then clear input of all the s receives ‘ ’ a d utputs f the th

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GATE QUESTION BANK

counter will be reset. When count = 1010, Y=1 and counter will be reset. Q3 Q2 Q1 Q0 1 0 1 0 1 1 0 0 1 1 1 0 9.

)

(

10.

So sequence is 0,1,2,3,4,5,4,3,2,1,0,1,2. 14.

[Ans. B] Let A1 A0 be the bits of number A and B1B0 be the bits of number B and let Y be the output A1 A0 B1 B0 Y 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 A1 A0 B1 B0 Y 1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 From the truth table we see that the u ber f ti es ‘ ’ bec es is

15.

[Ans. A]

[Ans. A] (

(

)

)

[Ans. B] f

t where N – no. of inverters and t - Propagation delay of each, so t sec 11.

[Ans. A] Since triggering in positive edge triggering & Q of pervious flip flop is input to next hence always down

12.

[Ans. C]

Digital Circuits

̅ Also the truth table ̅

J

K ̅

1

1

0

1

0

1

1

0

1 1 ̅

13.

[Ans. D] Initially Q=0 and count up ( ̅ =1) is active so it started counting up and when it reaches to 5 then decoder output at pin 5 becomes 0 and preset will be active and it will set Q and it will make the counter mode down and count becomes 4, then 3 then 2 then 1 then 0, as soon as it reaches 0, decoder output at pin 0 is low and clear is active and Q goes to 0 and ̅ = 1 so up is active a d it c u ts ……

next

= . . = A.CLK + Q

next = A.CLK + If CLK = 1 and A and B = 1

then

} No race around

If CLK = 1 and A = B = 0 } No race around

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GATE QUESTION BANK

Digital Circuits

Thus race around does not occur in the circuit 16.

[Ans. D] State table A

D

0

0

1

1

0

1

0

0

1

0

0

0

1

1

1

1

From State table A=O

Q=O

Q=1 A=O A=1

A=1 17.

[Ans. B] State table Present state

Next state

0 0 0 0 1 1 1 1 1 1 0 0 00, 01, 11, 10, 00 18.

1 1 0 0

0 1 1 0

1 1 0 0

[Ans. C] i c c freque cy z

z

5 i

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GATE QUESTION BANK

Digital Circuits

AD /DA Convertor ECE-2006 1. A 4-bit D/A converter is connected to a free-running 3-bit UP counter, as shown in the following figure. Which of the following waveforms will be observed at V 0?

3.

D3 D2 D1 D0

Q1 Q0 Clock 3-bit Counter

IDEAL Vo

1k

(C) – 3.125 V (D) – 6.250 V

ECE-2008 Statement for linked Answer Questions 4 and 5 In the following circuit, the comparator output is logic “1” if and is logic “0”otherwise. The D/A conversion is done as per the relation = ∑ 2 Volts, where (MSB), and (LSB) are the counter outputs. The counter starts from the clear state.

1k Q2

The voltage Vo is (A) – 0.781 V (B) – 1.562 V

D/A Convertor

In the figure shown above, the ground has been shown by the symbol (A)

4 bit D/A converter +5V

Binary to BCD

̅̅̅̅

(B) ̅̅̅̅̅

2 Digit LED Display

4bit Upcounter

= 6.2

Clock

(C) 4.

The stable reading of the LED displays is (A) 06 (C) 12 (B) 07 (D) 13

5.

The magnitude of the error between and at steady state in volts is (A) 0.2 (C) 0.5 (B) 0.3 (D) 1.0

(D)

ECE-2007 Statement for Linked Answer Questions 2 and 3 In the Digital-to-Analog converter circuit shown in the figure below, = 10 V and R = 10kΩ. R

R

R

I

2R

V R

2R

2R

2R

2R

R +

2.

The current I is (A) 31.25 μA (B) 62.5 μA

V0

ECE-2011 6. The output of a 3 – stage Johnson (twisted – ring) counter is fed to a digital – to – analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output is

(C) 125 μA (D) 250 μA

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GATE QUESTION BANK

(B) droop rate decreases and acquisition time increases (C) droop rate increases and acquisition time decreases (D) droop rate increases and acquisition time increases

D/A Converter D

Clock

D

D

Johnson Counter

(A)

Digital Circuits

EE-2006 1. A student has made a 3 – bit binary down counter and connected to the R – 2R ladder type DAC [Gain = ( 1 KΩ/2R)] as shown in figure to generate a staircase waveform. The output achieved is different as shown in figure. What could be the possible cause of this error?

(B)

2R

1kHz clock

R

R

R

2R

2R

2R

1kΩ +12V Vo 12V

Counter 10kΩ

7 6 (C)

5 4 3 2 1 0 ms

0 1 2 3 4 5 6 7

(A) The resistance values are incorrect (B) The counter is not working properly (C) The connection from the counter to DAC is not proper (D) The R and 2R resistances are interchanged

(D)

ECE-2014 7. For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then (A) droop rate decreases and acquisition time decreases

IN-2006 Common Data for Questions 1 and 2 An R-2R ladder type DAC is shown below. If a switch status is ‘0’ 0 is applied and if a switch status is ‘1’ 5 is applied to the corresponding terminal of the DAC.

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GATE QUESTION BANK R

R 2R

2R

2R

1.5V will produce a digital output of (A) 90H (C) 9BH (B) 96H (D) A0H

+

2

+ 5V

1.

What is the output voltage (V0) for the switch status S0=0,S1=1,S2=1? (A)

V

(B) 2.

V

(C) (D)

. .

Digital Circuits

V

IN - 2010 5. A 4-bit successive approximation type ADC has a full scale value of 15V. The sequence of the states, the SAR will traverse, for the conversion of an input of 8.15V is (A) ( )

V

What is the step size of the DAC? (A) 0.125 V (C) 0.625 V (B) 0.525 V (D) 0.75 V

IN-2007 3. The circuit shown in the figure below works as a 2-bit analog to digital converter for 0 ≤ ≤3 . 3V 0.5kΩ

(C)

tart conversion

tart conversion

1 0 0 0

1 0 0 0

1 1 0 0

0 1 0 0

1 0 1 0

0 0 1 0

1 0 0 1

0 0 0 1

1 0 0 0

0 0 0 0

End conversion

End conversion

(D)

tart conversion

tart conversion

1.0kΩ 1.0kΩ

Digital

1 0 0 0

1 0 0 0

Circuit

0 1 0 0

1 1 0 0

0 1 1 0

1 1 1 0

0 1 1 1

1 1 1 1

1 0 0 0

1 1 1 1

End conversion

End conversion

0.5kΩ

The MSB of the output , expressed as a Boolean function of the inputs , , , is given by (A) (C) (B) (D) + IN-2009 4. An 8- it ADC with 2’s complement output, has a nominal input range of 2V to +2V. It generates a digital code of 00H for an analog input in the range – 7.8125mV to +7.8125mV. An input of

IN-2014 6. A thermopile is constructed using 10 junctions of Chromel-Constantan (sensitivity 60μ /°C for each junction) connected in series. The output is fed to an amplifier having an infinite input impedance and a gain of 10. The output from the amplifier is acquired using a

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GATE QUESTION BANK

10-bit ADC, with reference voltage of 5 V. The resolution of this system in units of °C is _________ 7.

DAC output and input to be positive, the maximum error in conversion of the analog sample value is: Input Comparator ample

An N-bit ADC has an analog reference voltage V. Assuming zero mean and uniform distribution of the quantization error, the quantization noise power will be: (A) (B)

(

)

(

)

(C) (D)

(

Digital Circuits

tart of conversion ( C) Control logic

l l

R 2R adder DAC

)

Clock

l

l

p counter l

utpt uffer

l

√

l

8.

Reset l

l

The circuit in the figure represents a counter-based unipolar ADC. When SOC is asserted the counter is reset and clock is enabled so that the counter counts up and the DAC output grows. When the DAC output exceeds the input sample value, the comparator switches from logic 0 to logic 1, disabling the clock and enabling the output buffer by asserting EOC. Assuming all components to be ideal,

(A) (B) (C) (D)

th

th

Ena le End of Conversion l (E C)

directly proportional to inversely proportional to l independent of directly proportional to frequency

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GATE QUESTION BANK

Digital Circuits


2.

7.

[Ans. B]

[Ans. B] The i/p to D/A will be 0000, 0001, 0010, 0011, 1000, 1001, 1010, 1011. So O/P will be in B waveform

In a capacitor drop rate in given an we know i = C

dv 1 dt C From the above relation it is clear that if capacitor value increases then the drop rate decreases because of inversion relation. C Also = C = it ⇒ t = ⇒t c i From the above relation, it is clear that if capacitor value increases acquisition time also increases as of proportionality relation ⇒

[Ans. B] I 2

I V

R

2R

R

I 4

R

2R

I 8 R

I 16

2R

2R

2R

I 16

I 4

R V0

A

Due to virtual ground, node A can be considered as ground so = 10K So I = So I = 3.

EE 1.

[Ans. C] =

4.

= 1mA = 62.5 μA

I R[ 4

I ]= 16

[Ans. C] Sequence generated is 7→111 0→000 3→011 1→001 5→101 2→010 1→001 3→011 6→110 Instead of 4→100 2→010 5→101 4→100 6→110 0→000 7→111 Simply just by observation we can say that error is because of lack of proper connections

3.125

[Ans. D] =

2

4

volts

Counter will stop when VDAC> 6.2 volt So counter will stop when = 1101 o ED will display 1101 i.e CD ⇒ 13 5.

6.

[Ans. B] In steady state = 0.5 6.5V So error = 6.5 6.2 = 0.3V

0

2

dv dt

4= IN 1.

[Ans. B] At switch status =0 = =1

[Ans. A] For the Johnson counter sequence D D D 0 0 0 0 1 0 0 4 1 1 0 6 1 1 1 7 0 1 1 3 0 0 1 1 0 0 0 0

R

2R

2R

R

2R

2R

Thevenin s equivalent

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GATE QUESTION BANK

Digital Circuits

i.e 2V → 128 2R

2R

1.5V →

2R

2R ≡

5 2

2R 5

5.

[Ans. A] By characteristics of SAR ADC.

6.

[Ans. * ] Range 0.800 to 0.833 For 10 junctions the thermopile voltage will be 600 μ = 0.6 m When it measured through amplifier of gain 10, its voltage will be 6 mV. So, the thermopile sensitivity after amplification is 0.006 V/° C Which will be input to ADC of resolution = /2 1 ADC resolution = 5/1023=0.00488 In terms of temperature, the resolution is 0.00488/0.006=0.814

7.

[Ans. A]

8.

[Ans. A] Maximum error is equal to step size which is directly proportional to .

R ≡

∴ 2.

15 2

R

=

[Ans. C] Step size =

3.

4.

= = 0.625

[Ans. B] Truth table of ADC is 0 0 0 1 ⇒

0 0 1 1 =

0 1 1 1

0 0 1 1

0 1 0 1

128 = 96

It is 96 for 1.5 analog input, while is given an 96 = 2 complement of +96 = 2’s complement of 01100000 =10100000 = A0H

5

5

.

[Ans. D] 8 – it ADC o/p is in 2’s complement form i.e it represents 127 128 i/p voltage Range = 2 to 2

aximum Error =

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GATE QUESTION BANK

Digital Circuits

Semiconductor Memory ECE-2014 1. If WL is the Word Line and BL is the Bit Line, an SRAM cell is shown in

̅̅̅̅

̅̅̅̅

̅̅̅̅

̅̅̅̅

EE-2009 1. The increasing order of speed of data access for the following devices is i. Cache Memory ii. CDROM iii. Dynamic RAM iv. Processor Registers v. Magnetic Tape (A) (v), (ii), (iii), (iv), (i ) (B) (v), (ii), (iii), (i), (iv) (C) (ii), (i), (iii), (iv), (v) (D) (v), ( ii), (i) , (iii), (iv) IN-2011 1. An bit RAM is interfaced to an 8085 microprocessor. In a fully decoded Scheme if the address of the last memory location of this RAM is 4FFFH, the address of the first memory location of the RAM will be, (A) 1000 H (C) 3000 H (B) 2000 H (D) 4000 H

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GATE QUESTION BANK

Digital Circuits


EE 1.

IN 1.

[Ans. B] For an SRAM construction four MOSFETs are required (2-PMOS and 2-NMOS) with interchanged outputs connected to each CMOS inverter. So option (B) is correct.

[Ans. B] Processor registers has highest speed. Followed by cache memory then dynamic ram (slower than static ram because of refreshing required)

[Ans. C] Capacity of chip = last memory address – First memory address+1 = last memory address – Capacity of chip+1 = 4FFFH – 2000H+1 =3000H

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GATE QUESTION BANK

Digital Circuits

Microprocessors ECE-2006 1. An I/O peripheral device shown in Figure (b) below is to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H - D7 H, its chip-select (̅̅̅) should be connected to the output of the decoder shown in Figure (a) below. A2

LSB

A3

3 8 Decoder

A4

MSB

3.

In the circuit shown, the device connected to Y5 can have address in the range

o de ice hip e ec o decode

0 1 2 3 4 5 6 7

̅̅̅̅̅̅ ̅̅̅̅̅̅ ̅

(A) 2000 20FF (B) 2D00 2DFF ̅̅̅̅

A7 A6 ̅̅̅

Data ̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅

Fig. (a) I/O Peripheral

A1 A0

̅̅̅ Fig. (b)

(A) output 7 (B) output 5

(C) 2E00 2EFF (D) FD00 FDFF

(C) output 2 (D) output 0

ECE-2010 2. For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is 3000 MVI A, 45H 3002 MOV B, A 3003 STC 3004 CMC 3005 RAR 3006 XRA B (A) 00H (C) 67H (B) 45H (D) E7H

ECE-2011 4. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is MVI A, 07H RLC MOV B, A RLC RLC ADD B RRC (A) 8CH (B) 64H

(C) 23H (D) 15H

ECE-2014 5. For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data ( ) from an external device is shown in the figure. The instruction for correct data transfer is

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GATE QUESTION BANK o ecode

⁄

̅̅̅̅̅ ̅̅̅̅̅̅

̅̅̅̅̅

̅̅̅̅̅

(A) MVI A, F8H (B) IN F8H 6.

(C) OUT F8H (D) LDA F8F8H

n

mic op oce o execu e “ H” wi h a ing add e oca ion 1FFEH (STA copies the contents of the accumulator to the 16-bit address location). While the instruction is fetched and executed, the sequence of values written at the address pins is (A) 1FH, 1FH, 20H, 12H (B) 1FH, FEH, 1FH, FFH, 12H (C) 1FH, 1FH, 12H, 12H (D) 1FH, 1FH, 12H, 20H, 12H

EE-2006 1. In an 8085 microprocessor based system, it is desired to increment the contents of memory location whose address is available in (D, E) register pair and store the result in same location. The sequence of instructions is (A) XCH G (C) INXD INRM XCH G (B) XCH G (D) INRM INXH XCH G 2.

How many times DCR L instruction will be executed? (A) 255 (C) 65025 (B) 510 (D) 65279

e ice

aa u

igi a inpu

Digital Circuits

A software delay subroutine is written as given below: DELAY: MVI H, 255 D MVI L, 255 D LOOP: DCR L JNZ LOOP DCR H JNZ LOOP

EE-2008 3. The contents (in Hexadecimal) of some of the memory location in an 8085A based system are given below Address Contents .. .. 26FE 00 26FF 01 2700 02 2701 03 2702 04 .. .. The contents of stack pointer (SP), Program counter (PC) and (H, L) are 2700H, 2100H and 0000H respectively, when the following sequence of instruction are executed, 2100 H: DAD SP 2101 H: PCHL The contents of (SP) and (PC) at the end of execution will be (A) PC = 2102 H, SP = 2700 H (B) PC = 2700 H, SP = 2700 H (C) PC = 2800 H, SP = 26 FE H (D) PC = 2A02 H, SP= 2702 H EE-2009 4. In an 8085 microprocessor, the contents of the Accumulator, after the following instructions are executed will become XRA A MVI B, F0H SUB B (A) 01 H (C) F0 H (B) 0F H (D) 10 H EE-2010 5. hen a “ dd ” in uc ion i executed, the CPU carries out the following sequential operations internally: th

th

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GATE QUESTION BANK

Note: (R) means content of register R ((R)) means content of memory location pointed to by R PC means Program Counter SP means Stack Pointer (A) (SP) incremented (PC) Addr ((SP)) (PC) (B) (PC) Addr ((SP)) (PC) (SP) incremented (C) (PC) Addr (SP) incremented ((SP)) (PC) (D) ((SP)) (PC) (SP) incremented (PC) Addr EE-2011 6. A portion of the main program to call a subroutine SUB in an 8085 environment is given below. : : LXI D, DISP LP: CALL SUB : : It is desired that control be returned to LP + DISP + 3 when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are (A) POP D (C) POP H DAD H DAD D PUSH D PUSH H (B) POP H DAD D INX H INX H INX H PUSH H

(D) XTHL INXD INX D INX D XTHL

Digital Circuits

EE-2014 7. An output device is interfaced with 8-bit microprocessor 8085A. The interfacing circuit is shown in figure

ecode u pu

̅ ̅

u pu

o

e ice

̅̅̅̅̅

̅

The interfacing circuit makes use of 3 Line to 8 Line decoder having 3 enable ̅ ̅ The address of the device lines is (A) (C) (B) (D) 8.

In an 8085 microprocessor, the following program is executed Address location - Instruction 2000H XRA A 2001H MVI B, 04H 2003H MVI A, 03H 2005H RAR 2006H DCR B 2007H JNZ 2005 200AH HLT At the end of program, register A contains (A) 60H (C) 06H (B) 30H (D) 03H

9.

In 8085 microprocessor, the operation performed by the instruction LHLD is (A) H (B) H (C) H (D) H

IN-2006 1. An 8085 assembly language program is given as follows. The execution time of

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GATE QUESTION BANK

2.

Digital Circuits

each instruction is given against the instruction in terms of T-state. Instruction T-states MVI B, 0AH 7T LOOP: MVIC, 05H 7T DCR C 4T DCR B 4T JNZ LOOP 10T/7T The execution time of the program in terms of T-states is (A) 247 T (C) 254 T (B) 250 T (D) 257 T

4.

A memory mapped I/O device has an address of 00F0H. Which of the following 8085 instructions outputs the content of the accumulator to the I/O device? (A) LXI H, 00F0H MOV M, A (B) LXI H, 00F0H OUT M (C) LXI H, 00F0H OUT F0H (D) LXI H, 00F0H MOV A, M

IN-2008 5. A part of a program written for an 8085 microprocessor is shown below. When the program execution reaches LOOP2, the value of register C will be SUB A MOV C, A LOOP1: INR A DAA JC LOOP2 JNC LOOP1 LOOP2: NOP (A) 63H (C) 99H (B) 64H (D) 100H

IN-2007 3. 8-bi igned in ege in ’ comp emen form are read into the accumulator of an 8085 microprocessor from an I/O port using the following assembly language program segment with symbolic addresses. BEGIN: INPORT RAL JNC BEGIN RAR END: HLT This program (A) Halts upon reading a negative number (B) Halts upon reading a positive number (C) Halts upon reading a zero (D) Never halts

A snapshot of the address, data and control buses of an 8085 microprocessor executing program is given below: Address 2020H Data 24H Logic high IO/ ̅ ̅̅̅̅ Logic high ̅̅̅̅̅ Logic Low The assembly language instruction being executed is (A) IN 24H (C) OUT 24H (B) IN 20H (D) OUT 20H

6.

A 2k×8 bit RAM is interfaced to an 8-bit microprocessor. If the address of the first memory location in the RAM is 0800H, the address of the last memory location will be (A) 1000H (C) 4800H (B) 0FFFH (D) 47FFH

IN-2009 7. The following is an assembly language program for 8085 microprocessors Address Instruction Mnemonic Code 1000H 3E06 MVI A, 06H 1002H C6 70 ADI 70H 1004H 32 07 10 STA 1007H 1007H AF XRA A 1008H 76 HLT th

th

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GATE QUESTION BANK

When this program halts, the accumulator contains (A) 00H (C) 70H (B) 06H (D) 76H 8.

Consider a system consisting of a microprocessor, memory, and peripheral devices connected by a common bus. During DMA data transfer, the microprocessor (A) only reads from the bus (B) only writes to the bus (C) both reads from and writes to the bus (D) neither reads from nor writes to the bus

IN-2010 9. In an 8085 processor, the main program calls the subroutine SUB1 given below. When the program returns to the main program after executing SUB1, the value in the accumulator is Address Opcode Mnemonic 2000 3E 00 SUB1: MVIA,00h 2002 CD 05 20 CALL SUB2 2005 3C SUB2: INR A 2006 C9 RET (A) 00 (C) 02 (B) 01 (D) 03 10.

(A) 3000H (B) 4FFFH 11.

d

a a ̅̅̅

b b b b

The subroutine SBX given below is executed by an 8085 processor. The value in the accumulator immediately after the execution of the subroutine will be: SBX: MVI A, 99H ADI 11H MOV C, A RET (A) 00H (C) 99H (B) 11H (D) AAH

dd e

u

x og ammab e

x

n e up con o e

aa u

x

Assuming vectored interrupt, a correct sequence of operations when a single external interrupt (Ext INT1) is received will be : (A) x → → a a ead→ (B) x → → → a a ead (C) x → → → dd e Write (D) x → → aa ead→ dd e ie

d

̅̅̅

(C) AFFFH (D) C000H

IN-2014 12. A microprocessor accepts external interrupts (Ext INT) through a Programmable Interrupt Controller as shown in the figure.

A 8-bit DAC is interfaced with a microprocessor having 16 address lines (A0...A15) as shown in the adjoining figure. A possible valid address for this DAC is ine o ine ecode

Digital Circuits

na og ou pu

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th

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GATE QUESTION BANK

Digital Circuits


2.

0100 [Ans. B] I/O address range D4 – D7 H ⇒ xx So, and should be. So, decoder output should be 5.

3.

5.

⏟

0111

RLC

1110

MOV B, A

1110

RLC

1100

RLC

1000

ADD B A 0000 1110 + B 0011 1000

⏟

⏟

⏟

That is F8 F8 memory address is selected So LDA F8F8 [Load accumulator from the content of memory location F8F8H] should be the instruction for data transfer

[Ans. B] Address bits should be as follows:

[Ans. C] MVI A, 07 H

H

[Ans. D] In this figure the chip select and will be selected if the memory address will be

Content of register A = 45 B = 45 CY =1 CY =0 A = 22 ⊕

6.

(and for hey can be he o 2D00 – 2DFF (Range) 4.

→

[Ans. C] MVI A, 45 MOV B, A STC CMC RAR XRA B

0110

[Ans. A] STA 1234H is stored as follows 1FFE Opcode of STA 1FFF 34 H 2000 12H After this 1234H will be loaded in the address bus. So the correct sequence of values at are 1F, 1F, 20, 12

)

he content of ‘ ’ he content of ‘ ’ he content of ‘ ’ he content of ‘ ’ he content of ‘ ’

EE 1.

[Ans. A] XCHG (HL) (DE) INR M (HL) = (HL) + 1 So this sequence stores the address into HL pair and then increment the content of memory location specified by HL pair.

2.

[Ans. D] DCR L executed = 255 × 255 + 255 – 1 = 65279 times

3.

[Ans. B] Given, (SP) = 2700H (PC) = 2100 H (HL) = 0000H th

th

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GATE QUESTION BANK

2100 H : DAD SP (SP) + (HL) → (HL) 2700 H + 0000 H = 2700 H stored in HL pair 2101 H : PCHL : The content of HL are transferred to (PC) So now (PC) = 2700H and (SP) also unchanged (PC) = 2700 H (SP) = 2700 H 4.

5.

6.

[Ans. D] XRA A MVI B, F0H SUB B

Output line 2 is selected So ⇒

⏟

[Ans. A] H XRA A H MVI B, H H MVI A, H H RAR

⇒ ⇒ ⇒ – B will go to accumulator 00 – F0 = 10

SUB

⇒ ⏟

⏟

[Ans. D] First of all content of PC is loaded into stack. i.e. address of next instruction to be executed is loaded onto stack. i.e. SP is decremented then PC is loaded by address given in call instruction.

CALL

̅

⏟

8.

[Ans. C] Call take 3 address locations. RET always returns to LP + 3 location, this stored in SP. So to return to LP + DISP + 3 we have to add DISP to SP. POP H DAD D PUSH H Normal call operation shown.

Digital Circuits

→ clear accumulator → B= H → A=

H

→ Rotate accumulation to right A = 10000001 → Decrement B, B= → ump o when z

H H

DCR B INZ 2005 200 AH HLT After 3 rotation H 9.

[Ans. C] LHLD instruction loads the value at memory location specified by the immediate Value in H and L pair register Value at will be stored in L register value of will be stored in H register

IN 1.

[Ans. C]

RET

2. 7.

[Ans. A] Since I/O device is memory mapped I/O Memory related instructions will be used for data transfer

[Ans. B] To enable decoder

th

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GATE QUESTION BANK

H H → load the address of memory in HL pair → load the memory by content of A 3.

4.

5.

Digital Circuits

6.

[Ans. B] Starting address 0800H, so last address = 0800 + 7FF = 0FFFH.

7.

[Ans. A] Last instruction is XRA A, so accumulator contents will be 00

8.

[Ans. C] In DMA transfer of data between source and destination takes place without any involvement of microprocessor. During both in read/write mode.

9.

[Ans. B] SUB 1: MVI A, 00H → H CALL SUB 2 → program will shift to SUB 2 address location SUB 2: INR A →A = 00H + 01H=001H RET → returned to main program The contents of Accumulation after execution of the above SUB 2 is 01H

[Ans. D] Address 2020H: The contents of lower and higher address lines are same i.e. 20H. So, this indicates that the technique is I/O mapped I/O. Data 24H: This indicates that the content of accumulator is 24 H, which have to sent to I/O port. ̅ (Logic High): ̅ high indicates that this is input/output operation. ̅̅̅̅ (Logic High): ̅̅̅̅ logic high indicates that read operation is inactive. ̅̅̅̅̅ (Logic Low): ̅̅̅̅̅ logic low indicates that write operation is active. So, by observing all the operations the appropriate instruction is OUT20H.

10.

[Ans. A] To select 2-4 line decoder, A15 = 0 To select the DAC, b2 should be active, i.e. A14 = 0 and A13 = 1. Reset all address lines can be either 0 or 1. So, address can be 001x xxxx xxxx xxxx Out of four choices this is satisfied by 3000 H only.

11.

[Ans. D] Instructions MVI A, 99H ADI 11H MOV C,A

[Ans. A] Execution will go to Loop 2 when A contain 64H = , at that time C contains 63 H.

12.

[Ans. A] e.g. 01H positive number → After RAL CY Accumulator 0 0000 0010 or 00000011 JNC is TRUE and again IN instruction will load accumulator with same port address, and microprocessor will go into infinite loop. Negative number → after RAL CY Accumulator 1 0000 0001 or 00000000 JNC is false and microprocessor will come out from the loop. So, this program halts upon reading a negative number.

th

Content of register A = 99 A = 99 + 11 = AAH C = AAH

[Ans. B] The correct sequence is external interrupts occurs at PIC, then it is transferred to microprocessor, then interrupt is acknowledge and finally data is read. th

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GATE QUESTION BANK

EMT

Electromagnetic Field ECE - 2007 1. If C is a closed curve enclosing a surface S, then the magnetic field intensity ⃑⃑ , the

Which current distribution leads to this field? [Hint: The algebra is trivial in cylindrical co-ordinates.]

current destiny ⃑ and the electric flux destiny ⃑⃑ are related by (A) ∬ ⃑⃑

s

∮(

(B) ∫ ⃑⃑ ⃑⃑⃑ (C) ∯ ⃑⃑

s

) ⃑⃑⃑

⃑⃑

∯(

)

⃑⃑

∫(

(D) ∮ ⃑⃑ 2.

⃑⃑

⃑⃑

∬(

(A) ⃑

ŷ

(B) ⃑

ŷ

(C) ⃑

ŷ

(D) ⃑

ŷ

(

(C)

s

√

(

(

(

√

) r

(

) r

⃑

(

) r

) )

s

If a vector field ⃑ is related to another vector field ⃑ through ⃑ ∇×⃑ ,which of

5.

the following is true? Note: C and Sc refer to any closed contour and any surface whose boundary is C. (A) ∮ ⃑ . ⃑⃑⃑ ∫ ∫ ⃑⃑⃑ . ⃑⃑⃑⃑ (B) ∮ ⃑ . ⃑⃑⃑

∫ ∫ ⃑⃑⃑ . ⃑⃑⃑⃑

)

(C) ∮ ∇ × ⃑ . ⃑⃑⃑

∫ ∫ ∇. ⃑⃑⃑ . ⃑⃑⃑⃑

)

(D) ∮ ∇ × ⃑ . ⃑⃑⃑

∫ ∫ ⃑⃑⃑ . ⃑⃑⃑⃑

) √

(

r

(D)

is a constant) √

⃑

(B)

A plane wave of wavelength is travelling in a direction making an angle with positive x-axis and positive y-axis. The ⃑ Field of the plane wave can be represented as (

⃑

(A)

6.

Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y-z plane and parallel to the y-axis. The other wire is in the x-y plane and parallel to the x-axis. Which components of the resulting magnetic field are non-zero at the origin?

)

ECE - 2008 3. For static electric and magnetic fields in an inhomogeneous source-free, medium which of the following represents the corr ct form of two of Maxw ’s equations? (A) ∇. (C) ∇× ∇×B ∇×B (B) ∇. (D) ∇× ∇.B ∇.B

z

y

x

(A) (B) (C) (D)

ECE - 2009 4. A magnetic field in air is measured to be x y ⃑ B B ( ŷ x̂) x y x y

th

x, y, z components x, y components y, z components x, z components

th

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GATE QUESTION BANK

EMT

ECE - 2010 7.

If ⃑

x â ∮ ⃑ . shown in the figure is, xyâ

over the path

y

a

The magnetic field inside the hole is (A) uniform and depends only on d (B) uniform and depends only on b (C) uniform and depends on both b and d (D) non uniform

3

C

1

X

0 ⁄ √

(A) 0 (B) √

⁄ √

(C) 1 (D) √

ECE - 2013 11.

ECE - 2011 8. Consider a closed surface S surrounding a volume V. IF r is the position vector of a point inside S, with ̂ the unit normal on S, the value of the integral ∯ r. ̂ is (A) 3V (B) 5V

Consider a vector field ⃑ r . The closed loop line integral ∮ ⃑ can be expressed as (A) ∯ ∇ × ⃑ ⃑⃑⃑⃑s over the closed surface bounded by the loop (B) ∰ ∇ ⃑ v over the closed volume bounded by the loop (C) ∭ ∇ ⃑ v over the open volume bounded by the loop (D) ∬(∇ × ⃑ ) ⃑⃑⃑⃑s over the open surface

(C) 10V (D) 15V

bounded by the loop ECE - 2012 Statement for Linked Answer Questions 9 and 10 An infinitely long uniform solid wire of radius a carries a uniform dc current of density . 9.

The magnetic field at a distance r from the center of the wire is proportional to (A) r for r < a and 1/r2 for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a (D) 0 for r < a and 1/r2 for r > a

10.

A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

ECE - 2014 12. In spherical co-ordinates, let â â denote unit vectors along the , directions. r

si

cos

t

r â

m

And .

si cos t r â m r represent the electric and magnetic field components of the EM wave at large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at r m is _______. 13.

th

xâ yâ zâ a |r| r iv r ∇ r = ____________

If r th

th

th


GATE QUESTION BANK

14.

Given the vector cos x si y â si x cos y â where â â denote unit vectors along x,y directions, respectively. The magnitude of curl of A is __________

15.

The electric field (assumed to be onedimensional) between two points A and B is shown. Let a be the electrostatic potentials at A and B, respectively. The value of in Volts is_________

EE - 2007 Statement for Linked Answer Questions 2 &3 An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of an air gap length of 1mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and a ag i ucta c . μo 4 x -7 H/m) 2. The current in the inductor is (A) 18.08 A (C) 4.56 A (B) 9.04 A (D) 2.28 A 3.

The average force on the core to reduce the air gap will be (A) 832.29N (C) 3332.47N (B) 1666.22N (D) 6664.84 N

4.

The total reactance and total susceptance of a lossless overhead EHV line, operating at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity of wave propagation is × km/s, then the approximate length of the line is (A) 122 km (C) 222 Km (B) 172 km (D) 272 km

5.

A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume. What is the magnitude of the electric field intensity, E, at a distance r (0 < r < R) inside the sphere?

⁄cm

4 ⁄cm ⁄cm

μm

B

16.

Given ⃑ zâ xâ yâ . If S represents the portion of the sphere x y z =1 for z the ∫ ∇ × ⃑ . ⃑⃑⃑⃑s is ___________

17.

If ⃑

y

yz x̂

xz ŷ xyz ẑ is the electric field in a source free region, a valid expression for the electrostatic potential is (A) xy yz (C) y xyz (B) xy xyz (D) xy xyz

xy

EE - 2006 1. Which of the following statements holds for the divergence of electric and magnetic flux densities? (A) Both are zero (B) These are zero for static densities but non zero for time varying densities (C) It is zero for the electric flux density (D) It is zero for the magnetic flux density

EMT

6.

(A)

(C)

(B)

(D)

Divergence of the vector field V( x, y, z) = (x cos xy+ y) i + (y cos xy)j + (sin z2 + x2 + y2) k is (A) 2z cos z2 (B) sin xy + 2z cos z2 (C) x sin xy – cos z (D) None of these.

th

th

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GATE QUESTION BANK

EE - 2008 7. A coil of 300 turns is wound on a nonmagnetic core having a mean circumference of 300 mm and a crosssectional area of 300 mm2. The inductance of the coil corresponding to a magnetizing current of 3A will be (Given that μ = 4 x 10-7 H/m) (A) 37.68 μH (C) 37. 68 μH (B) 113.04 μH (D) 113. 04 μH 8.

9.

10.

A capacitor consists of two metal plates each 500 × 500 mm2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4mm thickness and a layer of paper of 2 mm thickness. The relative permittivities of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that = 8.85 x F/m) (A) 983.33 pF (C) 637.5 pF (B) 1475 pF (D) 9956.25 pF Two point charges = 10 μC and = 20μC are placed at coordinates (1,1,0) and ( 1, 1,0) respectively, the total electric flux passing through a plane z = 20 will be (A) 7.5 μC (C) 15.0 μC (B) 13.5 μC (D) 22.5 μC An extra high voltage transmission line of the length 300 km can be approximated by a lossless line having propagation constant . radians per km. Then the percentage ratio of the length to wavelength will given by (A) 24.24% (C) 19.05% (B) 12.12% (D) 6.06%

EMT

EE - 2013 12. The curl of the gradient of the scalar field defined by x y y z 4z x is (A) 4xya yza zxa (B) 4a a a (C) 4xy 4z a x yz a y

zx a

(D) 13.

The flux density at a point in space given by B 4xa ya a m . The value of constant k must be equal to (A) 2 (C) +0.5 (B) 0.5 (D) +2

14.

A dielectric slab with 500 mm x 500 mm cross-section is 0.4 m long. The slab is subjected to a uniform electric field of a a mm. The relative permittivity of the dielectric material is equal to 2. The value of constant is . × m . The energy stored in the dielectric in Joules is (C) .5 (A) . × (D) (B) . ×

EE - 2014 15. is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity , the expression for the modified capacitance is a

⁄

(A) (B) 16.

EE - 2012 11. The direction of vector A is radially outward from the origin, with | | r where r x y z and k is a constant. The value of n for which ∇ A=0 is (A) 2 (C) 1 (B) 2 (D) 0 th

(C) (D)

⁄

The following four vector fields are given in Cartesian co-ordinate system. The vector field which does not satisfy the property of magnetic flux density is (A) y a z a x a (B) z a x a y a (C) x a y a z a (D) y z a x z a x y a th

th


GATE QUESTION BANK

17.

A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates.

19.

A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R r is (A) 4 r (C) 4 (B) 4 r (D) 4

20.

A perfectly conducting metal plate is placed in x-y plane in a right handed coordinate system. A charge of √ columbs is placed at coordinate (0, 0, 2). is the permittivity of free space. Assume ̂ ̂ ̂ to be unit vectors along x, y and z axes respectively. At the coordinate √ √ ) the electric field vector ⃑

ot

If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio is (A) 1:4 (C) 3:2 (B) 2:3 (D) 4:1 18.

EMT

(Newtons/Columb) will be ⁄√ o um s y

The magnitude of magnetic flux density ⃑B at a point having normal distance d meters from an infinitely extended wire carrying current of

is

i

ucti g

√ √

u its .

x

An infinitely extended wire is laid along the x-axis and is carrying current of 4 A in the +ve x direction. Another infinitely extended wire is laid along the y-axis and is carrying 2 A current in the +ve y direction. μ is permeability of free space. Assume ̂ ̂ ̂ to be unit vectors along x, y and z axes respectively.

am s

rf ct y co m ta at

(C) ̂ (D) √ ̂

(A) √ ̂ ̂ (B)

μ

B

4

z

Assuming

right

coordinate system, magnetic field intensity, ⃑⃑ at coordinate (2,1,0) will be ̂w B

4

handed

r m

̂

m

̂

m m th

th

th


GATE QUESTION BANK

EMT


[Ans. D]

μ

*

B x

B +a y

∇× From stokes theorem .

∮ 2.

∬(

5.

). s

[Ans. B] rom sto

’s th or m

∮

[Ans. A]

∮∇×

s

z axis

iv

∇×

∮ ⃑ ⃑⃑⃑ x axis

∬ ⃑ ⃑⃑⃑⃑s

x axis

6.

[Ans. D] z

z axis

cos

√

(

√

) y

cos ⃑ ⃑ 3.

.

.

.

√

.

[Ans. D] rom Maxw ∇.B

â

x

â × â â â × â â Non-zero field components x and z

â

’s quatio 7.

[Ans. C] ⃑ xyâ

∇×

xâ

For static magnetic field,

yâ xâ

∇× 4.

⃑.

[Ans. C] ∇×B μ ∇×B μ a μ

*

yâ

xy x

x

y

y

3

a

||

a

1

S

R

P

Q

|| B

μ

x â

B a z

X

0

B

⁄ √

B a z B ( x

P-Q:

y=1 dy=0

∫ ⃑ . ⃑⃑⃑

B )a + y

⁄ √

∫

⁄ √

x x

⁄ √

x

|

⁄ √ ⁄ √

Q-R: th

th

th


GATE QUESTION BANK

∫ ⃑ . ⃑⃑⃑

11.

x

√

∫ (

√

)

y

4

[Ans. D] By curl stroke theorem ∮⃑ . c ∬ ∇ × ⃑ . s ov r o

R-S: Y=3 ,dy=0 ∫ ⃑ . ⃑⃑⃑

∫

⁄√

⁄ √

x x

12.

x | ⁄√

⁄ √

= (

∫ ⃑ . ⃑⃑⃑

â

√

)

y

∫ ⃑.

=1

9.

[Ans. D] Apply the divergence theorem

10.

a

r

a

r

a

r

a

r

a

si ∫

si

∫

.

∭ ∇. r

r

. .

and r is

13.

[Ans. C] r

â

â r si

∫

( ∇. r ∭ v the position vector)

si

r

. si â r Average power crossing

∫ ⃑ . ⃑⃑⃑

∫ ⃑ . ⃑⃑⃑

∯ r. ⃑ x

.

si

r

∫⃑

8.

m

si â m r So, average poynting vector is ⃑ (⃑ × ⃑⃑ )

∫ (

So, ∮ ⃑ .

si

r .

⃑⃑

x

√

∫ ⃑ . ⃑⃑⃑

)

surfac

[Ans. *] Range 55.4 to 55.6 In phasor notation ⃑

S-P: x=

EMT

r .ra r a

[Ans. *] Range 2.9 to 3.1 r̅ xâ yâ zâ r x y z r x r y r z x r y r z r r

x x

r

∇

r

r .∇

r

x

r

iv r . ∇

r

r

r . â r x

r

r r r [ . â x r

r ) x x r (r. ) z z r ( ) x (r.

[Ans. C] Magnetic held inside hole depends on radius of hole i.e b and also on the location of hole from center of the conductor ie d. Hole has uniform cross section magnetic field is uniform

x

(r.

r. [ th

r

th

r ) x

r.

x r ] r x

r r â â y r z r r â â ] y z

r x

y

(r.

r ) y

x ( ) r th


GATE QUESTION BANK

x x ) ( ) r r r x x ( ) r r r (r. ) x x

∫si

r. (

14.

15.

[Ans. *] Range 0.01 to 0.01 si x × cos y cos x × si y cur x y cos x × cos y cos x × cos y Magnitude of curl of A = 0

17.

×

x ⁄

×

x

(4 ×

x x)|

×

×

×

× × ×

16.

×

EE 1.

[Ans. *] Range 3.13 to 3.15 ̅ zâ xâ yâ â â â ∇× ̅

|

x z

â x y sphere

y x â

z

z y â

| 2.

[cos

cos

. ]

[Ans. D] ̅ y

. 4

yz x̂

xy xyz ẑ

xz ŷ

[Ans. D] Inductance ×

L=

represents a hemi-

×

×

×

= 0.32115H

r si cos r si si r cos

∫ ∇× ̅ . s

.

∫si

[Ans. D] ∇.B ∇. ot a ways Zero)

a x y z

si

si

∇v Hence tio ∇v [ y â xy z â yz â ] tio B [ y ∇v yz x̂ xy xz ŷ ] xyz ẑ tio [yz x̂ ∇v y xz ŷ xyz âz] tio [ y ∇v yz x̂] xy xz ŷ xyz ẑ Hence D option is correct

×

4×

∫

∫si

̅

.

∫

cos

∫ ∇×̅ . s

[Ans. *] Range . . ⁄cm ⁄cm ; A= ⁄cm 4 ⁄cm B= × 4 x × 4× x ∫

∫

EMT

I= x

y

z

=

×

r s ∫(â

â

× .

.

si â ) . s th

th

th


GATE QUESTION BANK

3.

[Ans. A]

9.

[Ans. C] Net flux will be half of the total charge,

Force = Energy = F= 4.

=

= 0.835J

.

. 4

10.

[Ans. D] m wav

√ ×

f

.4 .

.

u

√

(

×

u

Length = 3×

× .

×

×

×

× ×

.

11.

.

=185 km 5.

= 15µC

[Ans. B] v

EMT

[Ans. A] ⃑ r ̂ ∇. ⃑

×

m 4 4 .

. gth ) gth

.

4 4 .

r . r .

’i radius r =

ra ia s

gth

i wav

[Ans. A] harg

.

r

r

. r

r or ∇. ⃑ r r

6.

r

r

[Ans. A] iv rg c

12.

[Ans. D] cur [∇ sca ar]

13.

[Ans. A]

∇. =

∇. B

cos xy yx si xy cos xy y si xy cos z . z zcos z 7.

[Ans. B] L= =

×

×

x

4x

y

y

z

4 14.

[Ans. B]

15.

[Ans. A] Case (i)

× ×

= 113.09µH 8.

[Ans. B]

h

C=

. .

×

× ×

×

×

× × ×

×

×

= 4.4nF

×

. C = 1.467nF th

th

th


GATE QUESTION BANK

.h h

EMT

gio 4

.

Case(ii) h

(

i-

In any system the charge Q = CV is constant

ctric

.h

)

18.

[Ans. C]

h [

]

m t r am s

16.

[Ans. C] ̅ magnetic flux density [w If B ̅ ∇. B ̅ B â i. . if B B â B â ̅ ∇. B B B B Must x y z ̅ Option (A) : ∇. B ̅ Option (B) : ∇. B ̅ Option (C) : ∇. B x y ̅ Option (D) : ∇. B

sourc

m ]

m t r

4 am s sourc

qua to z ro ̅ B

̅ ̅ B B μ 4 m s ̂

̅ B

z

μ

̅ B 17.

[Ans. C]

μ

̅ B ̅ B

am s

4μ

μ

( ̂) ̂w

̂w

m ̂w

m

m

μ ̂ am

⃑⃑

m

̅ ̅ = cross product of Note: Direction of B current direction and radial drections respectively [Biot – savart’s aw] 4

gio

4

4 th

th

th


GATE QUESTION BANK

19.

[Ans. A]

EMT

× √ â 4 |r | |r̅ | √ (√ √ )

⃑ |r̅ | â

√ â |r | 4

√

√ √ √ √

The net flux leaving any closed surface is equal to charge enclosed in it Total electric flux leaving the co c tric s h r of ra ius ‘ ’ charg co tai y th m ta s h r of ra ius ‘r’ ot tia at a oi t o m ta ic s h r 4 4

(√ √

â

) √ √ √ √

√

⃑

*

√ √

√ √

√

̅

√ [ 4̂]

̅

̂

√

+

r wto ⁄ o um

r

iv

ot 4 r (flux ) leaving the sphere of radius ‘ ’ 4 r 20.

[Ans. B] Method of image: A perfect conducting plate acts like a mirror for the existing charge by the introduction of virtual charge opposite to the existing charge and equivalent distant from it z axis

X

√ ou om s

r̅ √ √

r̅

√

X

ou om s

⃑ ⃑

⃑

⃑ 4

imag

√ â |r |

√ â |r |

th

th

th


GATE QUESTION BANK

Measurement

Basics of Measurements and Error Analysis (Static & Dynamic Characteristics of Measuring Instrument) EE-2006 1. A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be and 30 minutes respectively. If the number of secondary turns is reduced by 1 the new ratio error (%) and phase error (min) will be respectively (A) 0. 0, 30 (C) (B) (D) 2.

A variable w is related to three other variables x, y, z as w = xy/z. The variables are measured with meters of accuracy 0.5% reading, % of full scale value and 1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainty in the measurement of will be (A) 0.5 % rdg (C) 6.7 % rdg (B) 5.5 % rdg (D) 7.0 % rdg

IN-2006 1. A certain quantity x is calculated from measured values of a, b, and c using the formula, x=k(

) Where k is a constant.

The maximum limiting error in each of the three measured quantities is Δ The maximum limiting error in x will be (A) 2(| | (B) (| |

| |

|) | | |) | |

(C) (

)

| |

(D) (

)

| |

IN-2007 2. Two sensors have measurement errors that are Gaussian distributed with zero means and variances and , respectively. The two sensor measurements and are combined to form the weighted average x= + (1 ≤ ≤ Assuming that the measurements errors of the two sensors are uncorrelated, the weighting factor that yields the smallest error variance of x is. (C) (A) (B)

3.

(D) 0.5

Consider a non-ideal voltage source whose output voltage is measured by a non-ideal voltmeter as shown below

Non-ideal Voltage source

ideal voltmeter

ideal voltage source

Non-ideal voltmeter

Let be the difference between and the measured voltage is function of (A) only (C) (B) only (D) IN-2009 4. A quantity x is calculated by using the formula x = (p q)/r, the measured values are p = 9, q = 6, r = 0.5. Assume that the measurement errors in p, q and r are independent. The absolute maximum error in the measurement of each of the three quantities is . The absolute maximum error in the calculated value of x is (A) (C) (B) (D) 6

th

th

th


GATE QUESTION BANK

5.

The response of a first order measurement system to a unit step input is 1 e , where t is in seconds. A ramp of 0.1units per second is given as the input to this system. The error in the measured value after transients have died down is (A) 0.02units (C) 0.2units (B) 0.1units (D) 1unit

IN-2010 6. The volume of a cylinder is computed from measurements of its height (h) and diameter (d). A set of several measurements of height has an average value of 0.2m and a standard deviation of 1%. The average value obtained for the diameter is 0.1m and the standard deviation is 1%. Assuming the errors in the measurements of height and diameter are uncorrelated, the standard deviation of the computed volume is (A) 1.00% (C) 2.23% (B) 1.73% (D) 2.41%

7.

IN-2013 9. Two ammeters A and A measure the same current and provide readings and , respectively. The ammeter errors can be characterized as independent zero mean Gaussian random variables of standard deviations and , respectively. The value of the current is computed as: The value of which gives the lowest standard deviation of I is (C) (A)

y

(D)

(B)

IN-2014 10. The resistance of a wire is given by the expression Where ρ is the resistivity Ω-meter), L is the length (meter) and D (meter) is the diameter of the wire. The error in measurement of each of the parameters ρ L and D is . Assuming that the errors are independent random variables, the percent error in measurement of R is ___________.

A measurement system with input x(t) and output y(t) is described by the differential equation 3

Measurement

.The

static sensitivity of the system is (A) 0.60 (C) 1.67 (B) 1.60 (D) 2.67

8.

The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurements more than 10.15 mm will be (A) 230 (C) 15 (B) 115 (D) 2

th

th

th


GATE QUESTION BANK

Measurement

Answer Keys & Explanations EE 1.

( [Ans. A]

Considering only positive values of error

Nominal ratio Primary current n (

)

i

)

=(

)

= 7%

% Ratio error = IN 1.

[Ans. A]

Actual ratio n trun ratio

Given X = K (

(

)

X=K.Y Applying log on both sides log X = log K + log Y log K = 0 ( K is constant) = log Y = 2 log

n ) )

)

Let Y =

ii

when no. of secondary is reduced by 1

(

(

6 A

Actual ratio =

n (

)

iii

= 2 *| |

|

|+

Dividing eq. (ii) and (iii)

(

2.

[Ans. B] For the case given, the smallest error is

3.

[Ans. C] Measured voltage

)

atio rror v Phase angle error

( ) v v

Reduction of one or two turns of the secondary winding, no doubt, reduces the ratio error, but it has no effect on the phase angle error.

2.

[Ans. D] Given, w =

v

(

learly

.

v v

f(

)

)

Taking log on both sides, log w = log x + log y – log z th

th

th


GATE QUESTION BANK

4.

5.

[Ans. C] p q p q

10. r r

[Ans. B] R(s) = e

s

lim

s

s

s

lim

[Ans. *] Range 2.3 to 2.5 ρL DD d L dρ DD d ρ dL DD d ρL dρ DDD d

√(

d ) dρ

ρ

√(

ρ ) ρ

(

s s lim

d

s s

s s s

Measurement

L

L ) L

(

(

D ) D

D

D ) D

√

6.

[Ans. A] Volume = r h h = 0.2, 2r=0.1 = 1% = 1%

7.

[Ans. B] dy i o y t y t dt p p Applying Laplace transform 3s Y(s) + 5Y(s) = 8X(s) Y(s) [3s+5] = 8X(s) s s s s s s

6 (

s)

s

8.

[Ans. B] By Gaussian distribution, the expected Measurement are 115.

9.

[Ans. A]

th

th

th


GATE QUESTION BANK

Measurement

Measurements of Basic Electrical Quantities 1 (Current Voltage, Resistance)

EE-2010 2. An ammeter has a current range of 0-5A, and its internal resistance is 0.2 . In order to change the range to 0-25A, we need to add a resistance of (A) 0.8 in series with the meter. (B) 1.0 in series with the meter. (C) 0.04 in parallel with the meter. (D) 0.05 in parallel with the meter. EE-2012 3. A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil (PMMC) meter connected across the same load reads. v(t)

I

(A)

12

(D)

6.

Suppose that resistors n are connected in parallel to give an equivalent resistor R. If resistors n have tolerance of 1% each, the equivalent resistor R for resistors = 300 and = 200 will have tolerance of (A) 0.5% (B) 1% (C) 1.2% (D) 2%

7.

Two ammeters X and Y have resistances of 1.2 and 1.5 respectively and they give full-scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by connecting shunts so as to give full scale deflection with 15 A. The ammeters along with shunts are connected in parallel and then placed in a circuit in which the total current flowing is 15A. The current in amperes indicated in ammeter X is ______________

20 time (ms)

(C) 8 V (D) 10 V

√

EE-2014 5. The dc current flowing in a circuit is measured by two ammeters, one PMMC and another Electrodynamometer type, connected in series. The PMMC meter contains 100 turns in the coil, the flux density in the air gap is 0.2 Wb/m , and the area of the coil is 80 mm . The electrodynamometer ammeter has a change in mutual inductance with respect to deflection of 0.5 mH/deg. The spring constants of both the meters are equal. The value of current, at which the deflections of the two meters are same, is__________

5V 10

(C) √

(B)

10V

0 5V (A) 4 V (B) 5 V

j

j1

EE-2006 1. A current of – 8 + 6 √ (sin + 300) A is passed through three meters. They are a centre zero PMMC meter, a true rms meter and a moving iron instrument. The respective readings (in A) will be (A) 8,6,10 (C) 8,10,10 (B) 8,6,8 (D) 8,2,2

EE-2013 4. Three moving iron type voltmeter are connected as shown. Voltmeter reading are n , as indicated. The correct relation among the voltmeter reading is

th

th

th


GATE QUESTION BANK

8.

Measurement 1.6V

An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150 V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150. If the wattmeter is used with 150 V voltage setting and 10 A current setting, the multiplying factor of the wattmeter is____________

100Ω 500Ω

1000Ω

Rs=100Ω Ig Eu

9.

A periodic waveform observed across a load is represented by 1 in () { 1 in 1 The measured value, using moving iron voltmeter connected across the load, is (A) √

(C)

(B) √

(D)

(A) 0.9910 V (B) 0.9991 V 3.

The voltmeter shown in the following figure has a sensitivity of 500 /V and a full scale of 100V. When connected in the circuit as shown, then meter reads 20 V. The value of Rx is + Rs

IN-2006 1. A 4 k, 0.02 W potentiometer is used in the circuit shown below. The minimum value of the resistance Rs in order to protect the potentiometer is

(A) 2.23k (B) 2.71k

k

Rx

(A) 75 k (B) 50 k

V

(C) 25 k (D) 10 k

An ac voltmeter uses the circuit shown below, where the PMMC meter has an internal resistance of 100  and requires a dc current of 1 mA for full scale deflection.

+

Rp

1

100 V

4. 15V

(C) 1.0012 V (D) 1.0048 V

Vo

(C) 3.82k (D) 8.92k

+ ~

2.

An unknown voltage source Eu (with negligible internal resistance) is connected to a potentiometer circuit as shown in the following figure. If the galvanometer current is 10𝜇A with the direction as indicated, then value of Eu is

AC input nput

+

PMMC meter

~

Assuming the diodes to be ideal, the value of Rs to obtain full scale deflection with 100V (ac rms) applied to the input terminal would be Rs (A) 80 k (C) 89.9 k (B) 89 k (D) 90 k

th

th

th


GATE QUESTION BANK

5.

A voltmeter has a range of [4 V, 20 V] and a resolution of 1 mV. The dynamic range is (A) 24 dB (C) 72 dB (B) 60 dB (D) 84 dB

9.

Measurement

The op-amp based circuit of a half wave rectifier electronic voltmeter shown below uses a PMMC ammeter with a full scale deflection (FSD) current of 1 mA n oil re i n e of 1k . The v lue of R that gives FSD for a sinusoidal input voltage of 100 mV (RMS) is

IN-2007 6. The figure shows a potentiometer of total resistance with a sliding contact.

Sinusoidal input

1k PMMC Ammeter

P + RL

R

Q

The resistance between the points P and Q of the potentiometer at the position of the contact shown is and the voltage

(A) (B)

ratio

Statement for Linked Answer Question for 10 and 11 In the wheatstone bridge shown below the galvanometer G has a current sensitivity of 1 mm a resistance of 2.5 k and a scale resolution of 1 mm. let be the minimum increase in R from its nominal value of 2 that can be detected by this bridge.

at this point is 0.5. If the ratio

=1, the ratio √

(A) (B) IN-2008 7. If a

is (C)

(D) 1+√

√

current

1+√

of

[

√

in(1

)

⁄ ) is √ o ( √ ]A passed through a true RMS ammeter, the meter reading will be (C) 12A (A) 6√ (D) √ 1 (B) √1 8.

A 2A full-scale PMMC type dc ammeter has a voltage drop of 100 mV at 2A. The meter can be converted into a 10A fullscale dc ammeter by connecting a (A) 1 . m re i or in p r llel wi h he meter (B) 1 . m re i or in erie wi h he meter (C) . m re i or in p r llel wi h he meter (D) . m re i or in erie wi h he meter

(C) 1 (D) 1

7.

.

k

k 1

k

10.

When R is 2 k (A) 6 V (B) 6.0024 V

11.

The value of (A) . (B) .

th

th

is (C) 6.0038 V (D) 6.005 V is approximately (C) . (D) 1

th


GATE QUESTION BANK

IN-2009 12. A galvanometer with internal resistance 1 n full-scale current 1mA is used to realize a dc voltmeter with a full scale range of 1V. The full scale range of this voltmeter can be extended to 10V by connecting an external resistance of value (A) 9 k (C) 1 k (B) 9.9 k (D) 11 k 13.

The dc potentiometer shown in the figure has a working current of 10mA with switch S open. Let Rg+R1=100 . The galvanometer G can only detect currents gre er h n 1 . The m ximum percentage error in the measurement of the unknown emf Ex as calculated from the slider position shown is closest to 3V

2m

3m

G

(A) 0.3 (B) 0.5

15.

v()

~

(A) 100 A (B) 70.7 A

(C) 63.7 A (D) 31.8 A

16.

The deflection angle of the pointer of an ideal moving iron ammeter is for 1.0 ampere dc current. If a current of 3sin(314t) amperes is passed through the ammeter then the deflection angle is (A) (C) (B) (D) 9

17.

The PMMC ammeter A in the adjoining figure has a range of 0 to 3mA. When switch is opened, the pointer of the ammeter swings to the 1mA mark, returns and settles at 0.9mA. The meter is

(C) 0.6 (D) 1.0

IN-2010 Common Data for Questions 14 & 15 A PMMC type ammeter has full scale current of 100 A and a coil resistance of 100 14.

The above PMMC meter is connected in the circuit shown in the adjoining figure. The op amp is ideal. The voltage vi(t) = 1.0sin314t V . Assuming the source impedance of vi (t) to be zero, the ammeter will indicate a current of

10k

S

R1

Measurement

A

1. k

1.8 V

The resistance required to convert the 100 A ammeter into a 1A full scale dc ammeter is (A) 10m in series with the meter (B) 10m in parallel with the meter (C) 1m in series with the meter (D) 1m in parallel with the meter

(A) critically damped and has resistance of 100 (B) critically damped and has resistance of 200 (C) under damped and has resistance of 100 (D) under damped and has resistance of 200 th

th

th

a coil a coil a coil a coil


GATE QUESTION BANK

IN-2011 18. In the Wheatstone bridge shown below, when the resistance in re e by 1 the current through the galvanometer is

1k

1k 10V

G 1k

S1

1k

Measurement

EE/IN-2012 20. An analog voltmeter uses external multiplier settings. With a multiplier setting of 20 k it reads 440 V and with a multiplier setting of 80 k, it reads 352 V. For a multiplier setting of 40 k the voltmeter reads (A) 371V (C) 394V (B) 383V (D) 406V

1k

(Consider the Thevenin equivalent resistance of the bridge in the calculations) (A) 1.25 µA (C) 12.5 µA (B) 2.5 µA (D) 25 µA IN-2012 19. A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil (PMMC) meter connected across the same load reads

IN-2013 21. The circuit below incorporates a permanent magnet coil milli-ammeter of range 1 mA having a series resistance of 10 k . Assuming constant diode forward re i n e of forw r io e rop of 0.7 V and infinite reverse diode drop of 1.7 V and infinite reverse diode resistance for each diode, the reading of the meter in mA is mA + 1 k

v(t) 10V 1 k

5V 0

10

12

20

z

time (ms)

5

(A)v 4V (B) 5V

(A) 0.45 (B) 0.5

(C) 8V (D) 10V

th

th

(C) 0.7 (D) 0.9

th


GATE QUESTION BANK

Measurement


5.

[Ans. *] Range 3.0 to 3.4 Electro dynamometer

[Ans. C] PMMC reads average value i,e d.c., so its reading is – 8 value. RMS meter will read rms value of current RMS = √

(

√ √

T k

T

[Ans. D] k By KCL, And By KVL, Voltage drop k n should be same ∴ (5)(0.2)=20( ) 1 = 20

o

across

i i i

 in parallel

.

. 

6.

i ( . )

i

or T

) = 10 A

M.I. also reads RMS value, so its reading is also 10 A 2.

i 1

w number of on u or i (i ) w i i i i ( . ) 1 k k 1 . 1 k . 1 k .

[Ans. B] 1 1 1

5A

25A

1

1

.

3.

. 1 1

[Ans. A] Area enclosed (1 )(1 )

(

)( )

1

( )( ) = 80

( . 1) (

1

1

)

. 1

Time period = 20 Average reading = 4.

7.

[Ans. *] Range 9.9 to 10.3 Ammeter X (1 1 1 1. . 1

[Ans. D]

1

Because voltage drop across inductor Because voltage drop across capacitor j – 1 Net voltage j j j Voltmeter reads magnitude of the voltage only

1

.1 )

m

1.

Ammeter Y th

th

th


GATE QUESTION BANK

(1

)

.

1.

.

IN 1.

.

Measurement

[Ans. B] 0.2 =

1

250 mA

I=

1

√

and I =

1.

√

.71 k .71 k 1

(x)

2. 1. 250 mA

h

(x)

[Ans. D]

1. 150 mA

h1

1.6 V

(1 )

oop

1

I 1

1

1. .

oop

1. (x) 8.

9.

. 1 1

1

Ig

.

. 7

1 .1

[Ans. 2] In LPF wattmeter, T on the moving system is small owing to low power factor even when the current and potential coils are fully excited. Also the errors introduced due to inductance of pressure coil tend to be large at low power factors. So for calculating multiplying factor for a low p.f. wattmeter, p.f. mentioned on the wattmeter should be taken into account. Therefore, Multiplying Factor = (Current range used*Voltage range used*p.f)/Power at FSD Given, Power at Full scale reading = 150 Current Range used = 10 A Voltage Range used = 150 V Power Factor = 0.2 1 1 . Therefore m 1

1 1 Applying KVL in Loop 1, we get 1.6 – 100I 500I – 1000 (I + Ig)=0 1.6 – 1000Ig=1600I 1.6 – 1 1 1 1. . 1 9.9 7 1 1 Now in loop 2, 1 1 ( ) 1

1 1 9.9 7 1 11 1 .99 7 . 11 .99 7 1. 7 1.

3.

√1

( ) √

√1

1

1

[Ans. B] Total resistance of voltmeter is )(1 =( ) k So equivalent circuit is

+

+ 80 V

[Ans. A] M.I instrument reads RMS value 1

1

100 V

√

+

I 100 k

20 V

k

20 V th

..

th

th


GATE QUESTION BANK

I=

= 0.4 mA (

∴

.9

.

∴

.

1

1

.

.9 1 .9 ( . )( ) ( .9 )(1 ) 1 , since diodes are ideal] 9.9

.

[

1

(

)

( 1

∴

)

√1

1 1

√

Here resistances cannot has negative value. 1

Therefore

√

[Ans. D] r nge (level )

e olu ilon (

)

7.

[Ans. C] √

1

1m n 1 .9 1 bits (as number of bit can never be fraction) dynamic range = 6 n = 84 dB 6.

)

[Ans. C] .

5.

k

.

)(

bu (Given) )( ) ∴( ∴ Divide above equation on both sides with

) = (0.8 – 0.4) = 0.4 mA

∴ 4.

(

= 0.8 mA

And

Measurement

√

)

√( √ )

( (

√ √

√

) )

(

√ √

)

= 12 A 8.

[Ans. A]

(

[Ans. A] Meter 10A

(

)

Given data

2A

P

1

Given drop across meter = 100 mV. = (10 – 2) = 8A ∴ By KVL (8)( ) = 100 mV ∴ 1 . m in p r llel wi h mme er

.

From circuit v

v (

v v

[Ans. A] 1

1 1

(

)(

1

(

(

)( )(

√

m

) (By the concept of virtual ground) ∴ m Also, it can be observed that ∴ VC = 45 mV poin ‘ ’ VC = 1 R

1 1

1

9.

)

) )

th

th

th


GATE QUESTION BANK

1 ∴ 10.

1 1

1

1 1

12.

[Ans. A] 1  1m So in order to measure 1 V (maximum) we have to connect a series resistance R1 such that 1 9  Now to increase voltage range to 10V, we have to add another external resistance R2 such that ( ) 1 1 1 1 9 1 9  9k

13.

[Ans. B] With switch open redrawing the circuit



[Ans. B] k

k

1 (

)

k

Bridge can detect a change in R if galvanometer deflections. If change in R is minimum, then minimum deflection in galvanometer should be 1 mm for detection as 1 mm is i ’ re olu ion. o for this 1𝜇 current must flow through it. i. e. 1 1 . Now in outer loop, ) 1 1 1 ( 1 1 1 1 1 1 1 1 . 1 1 .99 1 So, 1 ( ) ) 1 (1 1. 99 1 = 0.004 + 5.9984 = 6.0024 V

10 mA

+

+

5m

ol ge rop ro i 250 ×10mA = 2.5V So the voltage drop across 5m long resistance is (3 – 2.5) = 0.5V ∴ Resistance of 5 m long wire is .

11.

Measurement

[Ans. B] Writing KVL in inner loop we get, 1 . 1 1 . .997 .99 m Writing KVL in another inner loop we get ( ) ( ) . . . 9 ( ) ( .99 . ) 1 . .

=

∴ m orre pon o n he voltage drop across it is 20 × 10 m = 0.2V Redrawing the circuit with switch (s) closed 3V

10 mA

2m

3m

0.2V

2.5V +

0.3V + ( B

th

th

G

)

A

1

th


GATE QUESTION BANK

∴ should be 0.2V for the galvanometer to show null deflection. u given ‘ ’ n e e urren gre er than 10µA. so the potential difference across it should be (1 1 ) = 1mV ∴ Error is 1mV & true - value is 0.2V ∴ % error = 14.

1

.

17.

[Ans. D] Calculating the resistance

+

1.8K

1.8V

= 0.5

(1 1

15.

.

1 18.

( ) ) (100) = (1 100µ) ( .9999) 1 m in p r llel wi h me er.

[Ans. A] 1k

1k

10V

A

B 1.

1k

V(t) = 1 sin 314 t = 0.1m i(t) = (0.1m) sin (314 t) ∴ = 0.1m ( .

∴

)

I

63.7µA

G

19. .

∴ ⟹

√ ( .

√ )

A r

[Ans. D] Assuming spring control ∝ for d.c ∝ for a.c ∴

1k

Thevenin equivalent circuit across AB. ( . ) . 9 m (1 1 ) (1 1. 1 ) 1. . 9 m (1. r 1)1 1.

vg

16.

.

The meter swings between 1mA and 0.9 mA, so it is inherently under damped system.

)

[Ans. C] Here, the waveform seen by the ammeter is

Here,

1.62V

By KVL 1.8 – 1.62 = V V = 0.18V ∴

1 I = 1A By KVL,

of the meter

0.9 mA

[Ans. B]

(

Measurement

= 2.1213 A

9 th

[Ans. A] As PMMC meter reads only DC value or average value and average value is equal to re un er he urve To l ime th

th


GATE QUESTION BANK

(

1

1

1 )

(

)

(

Measurement

) ol 21.

20.

[Ans. D] or ul iplier e ing Meter resistance = x Total resistance = Voltmeter reading ( ) ∝ x

√ .7 7 m 1 PMMC measures the average value .7 7

k k

[Ans. A] Current in the inverting arm

x

.

k

m

V x

∝

1

or ul iplier e ing Meter resistance = x Total resistance ( ) k x Voltmeter reading ( )= 352 V k

V x

∝ ∝

x 1

( x

x x x) x

(

1 x k or mul iplier e ing Total Resistance ( )

x)(

)

x

k k

th

th

th


GATE QUESTION BANK

Measurement

Measurements of Basic Electrical Quantities 2 (Power and Energy, Instrument Transformers) EE-2006 1. A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5A p-p as shown in the figure. The period is 20 ms. The reading in W will be

EE-2010 4. A wattmeter is connected as shown in the figure. The wattmeter reads Current coil Potential coil

(A) (B) (C) (D)

0

0

(A) 0 W (B) 25 W

(C) 50 W (D) 100 W

EE-2009 2. The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be a 3-phase Balanced supply 400 Volts 50 Hz.

Z1 = (100 + j0) Ω

Z1 b

Z2 (A) 0 (B) 1600 Watt

3.

Z1

Z2

Z2 = (100 + j0) Ω

CC PC

C

(C) 800 Watt (D) 400 Watt

Wattmeter

~

Zero always Total power consumed by Power consumed by Power consumed by

and

EE-2011 5. Consider the following statements: (i) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (ii) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. (A) (i) is true but (ii) is false (B) (i) is false but (ii) is true (C) both (i) and (ii) are true (D) both (i) and (ii) are false EE-2013 6. The input impedance of the permanent moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the reading of the voltmeter in volts is 1k

~

The pressure coil of a dynamometer type wattmeter is (A) highly inductive (B) highly resistive (C) purely resistive (D) purely inductive

14.14 sin (314 t)V

(A) 4.46 (B) 3.15 th

th

100 k

Voltmeter

(C) 2.23 (D) 0 th


GATE QUESTION BANK

7.

A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as = 300 Ω . Other bridge resistances are = = = 300 Ω . The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage in mV is +

+

(A) 56.02 (B) 40.83

(C) 29.85 (D) 10.02

EE-2014 8. Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second. Then the load impedance angle in radians is (A) 12 (C) 6 (B) 8 (D) 3 9.

While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100 W and 250 W. The power factor of the load is__________

10.

In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance in microfarad is_________ 35 Ω (

) 105 Ω

01

Measurement

IN-2006 1. The two-wattmeter method is used to measure power in a 3-phase, 3-wire balanced inductive circuit. The line voltage and line current are 400 V and 10 A respectively. If the load pf is 0.866 lagging, then readings of the two wattmeters are (A) 6000 W and 0 W (B) 5000 W and 1000 W (C) 4500 W and 1500 W (D) 4000 W and 2000 W IN-2007 2. A dynamometer type wattmeter with a single scale marked for the smallest power range, has two current ranges, namely, 0-5 A and 0-10 A as well as two voltage ranges, namely, 0-150 V and 0-300 V. To carry out a load test on a 230V/115V, 1kVA, single phase transformer, the wattmeter is used on the high voltage side. The voltage and current ranges are chosen for maximum utilization of the scale. The multiplying factor to be used in this case is. (A) 0.5 (C) 2.0 (B) 1.0 (D) 4.0 IN-2008 3. A 230 V, 5A, 50 Hz single phase house service energy meter has a meter constant of 360 rev/KWhr. The meter takes 50s for making 51 revolutions of the disc when connected to a 10 kW, unity power factor load. The error in the reading of the meter is (A) 0% (C) 2.0% (B) +0.5% (D) +2.0% IN-2011 4. Power in a there phase star connected balanced inductive load is measured by two wattmeter method. The phase voltage and phase current are 230 V and 5 A, respectively. The power factor of the th

th

th


GATE QUESTION BANK

Measurement

load is 0.707. The reading and of the two wattmeters are (A) = 298 W and = 1111 W (B) = 516 W and = 1924 W (C) = 1220 W and = 1220 W (D) = 1111 W and = 516 W EE/IN-2012 5. For the circuit shown in the figure, the voltage and current expressions are v(t) = ( )+ (3 ) a d ( ) )+ = sin ( (3 - +) + (5 )

i(t) Load

+

wattmeter

V(t)

(A)

cos

(B)

[

(C)

[

(D)

[

+ + +

+

]

] ]

th

th

th


GATE QUESTION BANK

Measurement


2.

[Ans. A] If we consider both the waves, we can see positive power and negative power in each case is equal Net power = 0 W i.e. OW [Ans. *] Not Matching with IIT Keys Wattmeter reading = current through C.C voltage across PC cos (phase angle b/w V and I) =

4.

6.

[Ans. A]

=

10 ms

= 4 120 = = 400 120 = 4 +120 400 120 = 800 w

3.

field corresponding to this current. This field acts in opposition to the current coil field. Thus the resultant field is due to current l only. Hence the error caused by the pressure coil current flowing in the current coil is neutralized

240

[Ans. B] It is difficult to have purely resistive pressure coil. The pressure coil has a small value of inductance, due to which error occurs in wattmeter readings.

20 ms

14

14 14 100 = 14 101 14 = = 4 46

=

( PMMC reads average voltage)

7.

[Ans. C] During the normal condition

[Ans. D] Considering the wattmeter to be an ideal wattmeter and as PC is connected across . So option D.

300

300 +

300

300

[Ans. B] I+

Current coil

= 600 = 12

Compensating coil

Load

Supply

5.

The current coil carries a current of l + and produces a filed corresponding to this current. The compensating coil is connected in series with the pressure coil circuit and is made as nearly as possible identical and coincident with the current coil. It is so connected that it opposes the field of the current coil. The compensating coil carries a current and produces a

8.

20

=

= 6 02985

=

6 = 0 02985 = 29 85

[Ans. C] (30

= √3

(30 + )

= √3 a

)

= √3

(

)

(

)

=2 a th

= th

1 √3

=

6 th


GATE QUESTION BANK

9.

10.

IN 1.

3.

[Ans. *] Range 0.78 to 0.82 ( ) a = √3 = √3 ( + ) √3 3 a = 7 = = 0 803

a

*√3 (

[Ans. D]

150 350

10 kW

Total revolution =

= 4 kW and 2.

100

4.

[Ans. A] For a balanced lagging load, connected in a star network. The two wattmeter readings are: (30 ) = (30 + ) = Given the load is balanced, hence line voltage is equal to phase voltage. Similarly for line and phase currents. Given, = 0 707 = 45 (30 45) = 230 5 = 1111 (30 + ) = (75 ) = 230 5 = 298

5.

[Ans. C] ()= + ()= ( Average power

0.866

)+=

0 866 = 30

= 2 kW =

= 2kW

1 ∫ 2

3 )+

d(

(3

)

)

The product of different frequency terms have zero average value. 1 1 = + 2 2 1 ( )+ = [ ( )] 2 = + +

[Ans. C] The wattmeter is connected towards H.V. side The rated current I =

360 = 50 rev

= 2%

= =

⁄

=

% error =

[Ans. 0.3] At balance = 1 1 = = j j0 1 = 35 = 105 1 1 105 = 35 j j0 1 105 0 1 = = 03 35

[Ans. D] Total power = √3 = √3 400 10 = 6kW = + and

Measurement

= 4 34

The suitable range is 0.5A and 300V wattmeter The load is connected across L.V. side Therefore the rated current 1 = = 8 69 115 ( ) =2 The multiplying factor =

= 0 a 5 harmonic does not contribute power as it is not present in V(t) 1 = [ + ] 2

th

th

th


GATE QUESTION BANK

Measurement

Electronic Measuring Instruments 1 (Analog, Digital Meters & Bridges, ADC type DVM) EE-2006 1. R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2. The source voltage is applied across R1 and R3. Under balanced conditions which one of the following is true? (A) (B) (C) (D) EE-2007 2. A bridge circuit is shown in the figure below. Which one of the sequences given below is most suitable for balancing the bridge? j

4.

Two 8-bit ADCs, one of single slope integrating type and other of successive approximation type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5V, the approximate time taken by the ADCs will respectively, be (A) (C) , (B)

,

(D)

,

EE-2009 5. An average-reading digital multimeter reads 10V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read. (C) √ (A) √ (D) 10√ (B) √

j

(A) (B) (C) (D)

~ First adjust R4 and then adjust R1 First adjust R2 and then adjust R3 First adjust R2 and then adjust R4 First adjust R4 and then adjust R2

EE-2010 6. he Maxwell’s bridge shown in the figure is at balance. The parameters of the inductive coil are jω

EE-2008 3. The ac Bridge shown in the figure is used to measure the impedance Z. If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be

j ω

~ (A) (B) (C) (D)

B . 98μF Oscillator 5 A

~ D

C

15.91 mH

Z

j

D

(A) (260+j0) (B) j

(C) (D)

j j th

th

th


GATE QUESTION BANK

EE-2011

7.

A 4

digit DMM has the error

IN-2006 1. Consider the AC bridge shown below. If ω

specification as: 0.2% of reading +10 counts. If a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can be expected in the reading is (A) ± 0.1% (C) ± 0.3% (B) ± 0.2% (D) ± 0.4%

8.

Measurement

⁄

and

(C) low Q inductor (D) lossy capacitor

(C)

(A) √

(D)

IN-2007 2. Consider the AC bridge shown in the figure below, with A, L and C having positive finite values.

~

EE-2014 9. The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is ____________

sinωt

.5

j

1/j

j

j

| |

~

(B)

(A) low resistance (B) high resistance

then ration |

is approximately equal to

The bridge circuit shown in the figure below is used for the measurement of an unknown element . The bridge circuit is best suited when is a

~

|

.

(A) (B)

if ω if

(C)

if

(D)

ω

√

cannot be made zero

sin ωt

10.

IN-2008 3. If the ac bridge circuit shown below is balanced the elements Z can be a

The saw-tooth voltage waveform shown in the figure is fed to a moving iron voltmeter. Its reading would be close to ______________

R Detector

~

D Z

(A) (B) (C) (D)

t ms

ms

th

Pure capacitor Pure inductor R-L series combination R-L parallel combination th

th


GATE QUESTION BANK

4.

A

digit, 200 mV full scale DVM has an

accuracy specification of .5 of reading plus 5 counts. When the meter reads 100 mV, the voltage being measured as (A) Any value between 99.5 mV and 100.5 mV (B) Any value between 99.0 mV and 101.0 mV (C) exactly 99.5 mV (D) exactly 100 mV

Measurement

IN-2014 6. The resistance and inductance of an inductive coil are measured using an AC bridge as shown in the figure. The bridge is to be balanced by varying the impedance z coil

EE/IN-2012 5. The bridge method commonly used for finding mutual inductance is (A) Heaviside Campbell bridge (B) Schering bridge (C) De Sauty bridge (D) Wien bridge

For obtaining balance z should consist of element(s): (A) R and C (C) L and C (B) R and L (D) Only C


2.

[Ans. C]

[Ans. B]

x

R1 i1 i2 R2

ω and x

R3

z z z

R4

z

jx

(

)

i3 D i4

ω

jx

jω

j ω Under balanced condition z z z z jω jx

Battery

Adjustments are made in various arms of the bridge so that the voltage across the detector is zero and hence no current flows through it, when no current flows through detector the bridge is said to be balanced. At balance condition

j [ω

] ω Equating real and imaginary terms, we obtain

ω

ω Solving above equations, we get ω

th

th

and

th


GATE QUESTION BANK

ω

4.

[Ans. B] Single slope integrating type ADC utilizes digital counter techniques to measure time required for a voltage ramp to rise from zero to the input voltage. If conversion time for input voltage 5V = TA So, conversion time for input voltage 2.5 V = TA /2 Conversion time in successive type ADC does not depend on input voltage. So, conversion time for input voltage 2.5 V is also TB.

5.

[Ans. A] For triangular wave

ω Q factor of the coil Q = Therefore

. ( )

and ( )

For a value of a greater than 10, the term (1/Q)2 will be smaller than 1/1000 and can be neglected Therefore eq. (1) and (2) reduces to ω . R4 appears only in eq. (4) and R2 appears in both eq. (3) &(4) So first R2 is adjusted and then R4 is adjusted.

Avg value = RMS value =

[Ans. A] 5

~

√

∴

B C= . 98 μF

Oscillator

3.

Measurement

RMS value =

√

√

RBC=300

6. D

A

[Ans. A]

C

L=15.1 mH

RAD

R+jω

Z

300

D

j ω

5

~ jω

jω .

j jω j

5.9

‖(

)

‖(

)

At balance

j At balance

jω

(

)

jω jω jω jω Equating real and imaginary terms, j ω

th

ω

th

th


GATE QUESTION BANK

7.

[Ans. C]

9.

Measurement


digit i

.5

(

i

j

√

So, maximum count with 9999

1 count =

] )

√

digit display

Full scale reading = 200 V

9

['i' is current through each branch which are same ]

No. of full digits in case of

digits display

√ 9

[

√

j

.

√ 10.

j

√

[Ans. *] Range 56 to59

E1 = Error corresponding to 10 counts .

9999

eading E2 = Error corresponding to 0.2 % of reading E2 = .

ms

.

RMS value of Saw tooth waveform is

Total error =

. . = 0.3 V %error in reading of 100V . . 8.

[Ans. C] The bridge is Maxwell inductance capacitance bridge. Element is an inductor Inductance = Lx effective resistance of the inductor = Rx ω ω i

ms t

Moving iron meter reads RMS value only

Meter reads

√

√

= 57.73 V IN 1.

[Ans. D] a C

R

R

The bridge is limited to measurement of low Q inductor (1 10, the bridge is unsuitable. This a Maxwell inductance capacitance bridge, so we can Say directly that it is used for low Q - value

C

C b

ω ω ω ω ω

ω

th

th

th


GATE QUESTION BANK

ω

ω

ω [

3.

[Ans. A] At balance,

ω ω

ω

ω

ω

ω

ω

]

(

ω ω ω ω

)( (

(

)

)

j

) j ω et j Comparing the real and imaginary parts we get

ω ω ω

Measurement

(

X=0

)

and

∴

(

j ω

)

j

ω ω is purely capacitive.

. ∴ It can be neglected in comparison to 2

4.

[Ans. B] Error on reading 100 mV m

.

.5 m

Also, error due to 5 counts 2.

[Ans. D] For ( (

) jω j ) ω

jω jω

j ω j ω Comparing the imaginary parts as real parts are already equal ω ω

5m .5 m 999 Hence, total error = 0.5 + 0.5 = 1.0 mV ∴ Meter will read between (100 – 1) mV and (100+1) mV. ie. Between (99.0 mV and 101 mV)

bridge must be balanced

5.

[Ans. A] Heaviside campbell bridge method is commonly used for finding mutual inductance.

6.

[Ans. B] At balanced condition,

ω √

√ ∴ frequency is coming to be imaginary which is impossible, hence, cannot be made zero.

jω jω ∴

Should be combination of R and L his bridge is known as Maxwell’s inductance bridge so we can directly say the should be the combination of ‘R’ and ‘L’

th

th

th


GATE QUESTION BANK

Measurement

Electronic Measuring Instruments 2 (C.R.O, RF Meters, Special Meters, Q meter) EE-2006 1. The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5V p-p square wave calibration pulse to channel 1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel 2 (lower trace) of the scope. If the time/div and V/div on both channels are the same, the amplitude (p-p) and period of the unknown signal are respectively

(A) 5 V, 1 ms (B) 5 V, 2 ms

A

B

h

L

G C

(A) A, B, C, A (B) A, B, C, B

Y

X

(C) 7.5 V, 2 ms (D) 10 V, 1 ms

EE-2007 2. The probes of a nonisolated, two channel oscilloscope are clipped to points A, B and C in the circuit of the adjacent figure. is a square wave of a suitable low frequency. The display on h and h are as shown on the right. Then the “Signal” and “Ground” probes S G and S G of h and h respectively are connected to points:

R

EE-2008 3. Two sinusoidal signals ( t) = A sin t and q( t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen is shown below: The signal q( t)will be represented as

h

(A) q( (B) q(

t) = A sin 2t)= A sin

2t,

2

=2 2t, 2 =

(C) q( (D) q(

2t)=

2t,

2=

2t,

2=

A cos 2t)= A cos

2

1

1

EE-2009 4. The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal (B) The amplitudes of the signals are very close but not equal (C) The signals are sinusoidal with their frequencies very close but not equal (D) There is a constant but small phase difference between the signals

(C) C, B, A, B (D) B, A, B, C

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GATE QUESTION BANK

EE-2011 5. A dual trace oscilloscope is set to operate in the alternate mode. The control input of the multiplexer used in the y-circuit is fed with a signal having a frequency equal to (A) the highest frequency that the multiplexer can operate properly (B) twice the frequency of the time base (sweep) oscillator (C) the frequency of the time base (sweep) oscillator (D) half the frequency of the time base (sweep) oscillator

The two signals S and S , shown in figure, are applied to Y and X deflection plates of an oscilloscope.

IN-2007 1. The linear sweep for the time base in an oscilloscope has deviation from its nominal waveform. The nominal (dashed line) and actual (solid line) sweep waveforms are shown in the following figure. Normal Sweep start time

S T

T

t

Retrace

Actual

time

T

7.

(C)

S

T

T

1.1T 1.15T 1.3T

EE-2014 6. In an oscilloscope screen, linear sweep is applied at the (A) vertical axis (B) horizontal axis (C) origin (D) both horizontal and vertical axis

Measurement

A 5V p-p sine wave with a frequency of 1 kHz will be measured on the oscilloscope as a sine wave with (A) 4.45 V p-p and 1 kHz frequency (B) 5 V p-p and 1kHz frequency (C) 5 V p-p and 1.1 kHz frequency (D) 5 V p-p and 1.15 kHz frequency

t

The waveform displayed on the screen is

2.

Two signals of peak- to-peak voltages 5 V and 2V are being fed to Channel 1 and Channel 2, respectively, of an oscilloscope with a single time base. The vertical sensitivity of both channels is 1 V/division. The two sine waves have identical frequency and phase. The trigger is on manual mode the triggers at a level of +1.25 V on Channel 1, as shown in the figure below:

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GATE QUESTION BANK

Which of the following figures correctly depicts the trace seen in channel 2?

Measurement

IN-2008 3. The x and y sensitivities of an analog oscilloscope are set as 2 ms/cm and 1V/cm respectively. The trigger is set at 0V with negative slope. An input of cos 00π t 300) V is fed to the y input of the oscilloscope. The waveform seen on the oscilloscope will be

(A)

(B)

(P) (C)

(D) (Q) IN-2009 4. The figure shows a periodic waveform to be displayed on a CRO. A trigger setting which ensures a stationary display is 2.5 2 1.5 1 0.5

(R)

0

-0.5 -1 -1.5 -2 -2.5

(A) (B) (C) (D)

(S) (A) P (B) Q

level: 0.2V, slope: ve level: 0.5V, slope: ve level: 0.2V, slope: +ve level: 1.8V, slope: ve

(C) R (D) S

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GATE QUESTION BANK

5.

The input impedance of CRO is equivalent to a 1M resistance in parallel with a 45pF capacitance. It is used with a compensated 10-1 attenuation probe. The effective input capacitance at the probe tip is (A) 4.5pF (C) 45pF (B) 5pF (D) 450pF

IN-2010 6. In an analog single channel cathode ray oscilloscope (CRO), the x and y sensitivities are set as 1ms/div. and 1V/div. respectively. The y-input is connected to a voltage signal 4 cos (200πt 45º) V. The trigger source is internal, level chosen is zero and the slope is positive. The display seen on the CRO screen is (A)

Measurement

IN-2011 7. A transfer characteristics of the circuit drawn below is observed on an oscilloscope used in XY mode. The display on the oscilloscope is shown in the right hand side. is connected to the X input with a setting of 0.5 V/div, and is connected to the Y input with a setting of 2 V/div. The beam is positioned at the origin when is zero.

sin 0πt

Assuming that the op amp is ideal and zener diodes have forward biased voltage drop of 0.7 V, the value of reverse break – down voltage of and are respectively (A) 3.3 V and 5.3 V (B) 4.7 V and 6.7 V (C) 6.7 V and 4.7 V (D) 5.3 V and 3.3 V

(B)

(C)

(D)

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GATE QUESTION BANK

Measurement


[Ans. C] Peak-peak (p – p) division of upper trace voltage = 2 value of (p – p) voltage = 5 V .

division

0

Now it will be same for unknown voltage (p – p) division of unknown voltage = 3 p – p voltage = 3 2.5 = 7.5 V Frequency of upper trace = 1 kHz Time period =

So, S is connected to point A and G is connected to point B. Voltage across inductor = d [ dt

= 1 ms

Division of x – axis (upper) = 4 Division of x – axis (lower) = 8 Period of unknown signal = 2 ms. 2.

t

t

So, S is connected to point C. and G is connected to point B. L

i(t)

3.

[Ans. D] Here p( t) = A sin t y line cuts 4 times the Lissagous patter x line cuts 2 times the Lissagous patter y

C

s s

s s (s

)

(

s I(t)

( (

)]

t

B

s

e

e

[Ans. B] Square wave is of low frequency. So, it can be assumed that time during which the wave form are displayed on the screen, the voltage across R and L is . A

(

x )

s

s

s

s

x

) e

)

Voltage across resistance i t e and q( t) will lead P( t) by 90 as trace is a circle q ( t) = A sin ( t + 90 ) q ( t) = A cos t. th

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GATE QUESTION BANK

4.

5.

[Ans. C] Because of phase difference only figure change from ellipse to circle and back to ellipse [Ans. D] Alternate mode is used to display two wave forms simultaneously by single CRT. As fluorescent material stores light for some time and eye sensing time is 20 ms. By using multiplexer alternatively two waves are connected to y plates. In this mode the frequency of control signal to multiplexer is equal to half off X – time base generator. In one sweep displays 1st waveform and in the half record sweep displays 2nd waveform connected to Y – plates. For low frequency signals multiplexer is switches at high frequency signals multiplexer is switches at high frequency that the multiplexer can operate. This is called chopping mode.

~

X0

time base

Y

X1

S

xy node y

x

Vector sum (in x y) Points A B C D

[Ans. B]

7.

[Ans. A] S

T

y 0 1 0

y 1 1 0

√x 1 2 0 2

y

Φ tan

( )

0 45° 0 225°

[Ans. C] It is clear from the figure that there is no change in amplitude, So it will show 5 V peak. ⇒ Also amplitude is function of Y – plates so no change. ⇒ But here slope during sweep will change and that will responsible for change frequency. ⇒ Slope is degraded by a factor 1.1, so frequency will be multiplied by 1.1 (as f ∝ 1/T). So f = 1 . . kHz.

S

6.

t

0

IN 1.

S

Measurement

t

2.

[Ans. B] As the two channel signals have same frequency, phase and the trigger level is set to 1.25 V which is below the peak – peak voltage of input signals (5V and 2V). So an identical trace appears on channel 2 also th

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GATE QUESTION BANK

3.

[Ans. A] X- sensitivity = 2 msec/ cm = Sx Y-sensitivity = 1 V/cm = Sy For input peak voltage, =4V So length on Y-axis from peak to peak. l

S

6.

[Ans. A] Vy(t) = 4 cos is

c

0 sec T/2 = 0 sec Now half cycle on X- axis correspond to length, T 0 sec l c S sec c This is shown in figure (A) only. [Hint:- Negative slope is given]

5.

[Ans. C] Hint: The signal is given with rising edge so the slope will be +ve and only option (C) satisfies this criteria.

set

00πt as

is set as

.

Internal triggering is chosen trigger voltage level is 0V & Slope is positive screen dimensions are 10div 8div Number of cycles of signal displayed f T 00 z 0div s div cycle The test signal is used as triggering signal since internal triggering is chosen. As the trigger voltage level is 0V, the signal will be displayed from 0V onwards and from rising side since trigger slope, is positive. Therefore option is ‘A’

⇒T

4.

Measurement

7.

[Ans. D] When is +ve [2 Division × 2V/division = 4 V] 0

[Ans. A] C1 C R

hen is ve is reverse bias

Probe

is forward bias and

0. 3.3 When

So effective capacitance

is negative

.

[3 Division × 2 V/Division = +6V] 0 0. p But effective input capacitance at probe tip

.3

. p

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GATE QUESTION BANK

Power Systems

Transmission and Distribution EE-2006 1. The concept of an electrically short, medium and long line is primarily based on the (A) nominal voltage of the line (B) physical length of the line (C) wavelength of the line (D) power transmitted over the line 2.

3.

4.

An HVDC link consists of rectifier, inverter transmission line and other equipments. Which one of the following is true for this link? (A) The transmission line produces/supplies reactive power (B) The rectifier consumes reactive power and the inverter supplies reactive power from/to the respective connected AC systems (C) Rectifier supplies reactive power and the inverter consumes reactive power to/from the respective connected AC systems (D) Both the converters (rectifier and inverter) consume reactive power from the respective connected AC systems The A, B, C, D constants of a 220 kV line are: A = D= 0.94 , B = 130 . C=0.001 If the sending end voltage of the line for a given load delivered at nominal voltage is 240 kV, the % voltage regulation of the line is (A) 5 (C) 16 (B) 9 (D) 21

2.5m and 3m respectively. The voltage (volt/km) induced in the telephone circuit, due to 50 Hz current of 100 amps in the power circuit is (A) 4.81 (B) 3.56 (C) 2.29 (D) 1.27 A generator is connected through a 20 MVA 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating, A 0.72 p.u. load, evaluated on load side transformer ratings as base values, is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the values of the load ( in per unit) in generator circuit will be (A) 36 (B) 1.44 (C) 0.72 (D) 0.18

5.

EE-2007 6. Consider the transformer connections in a part of a power system shown in the figure. The nature of transformer connections and phase shifts are indicated for all but one transformer. Which of the following connections, and the corresponding phase shift θ, should be used for the transformer between A and B?

~

Y

15kV

A

B θ

Y 400kV

Y

220kV

Autotransformer

A single phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common tower. The shortest and longest distances between the phase and telephone conductors are

(A) (B) (C) (D)

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Star – Star (θ = ) Star – Delta (θ = ) Delta – Star (θ = ) Star – Zigzag (θ = )

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GATE QUESTION BANK 7.

Two regional systems, each having several synchronous generators and loads are interconnected by an ac line and a HVDC link as shown in the figure. Which of the following statements is true in the steady state:

HVDC link Region 1

Region 2 AC line

(A) Both regions need not have the same frequency (B) The total power flow between the regions ( + ) can be changed by controlling the HVDC converters alone (C) The power sharing between the ac line and the HVDC link can be changed by controlling the HVDC converters alone. (D) The direction of power flow in the HVDC link ( ) cannot be reversed. 8.

Consider a bundled conductor of an overhead line, consisting of three identical sub-conductors placed at the corners of an equilateral triangle as shown in the figure. If we neglect the charges on the other phase conductors and ground, and assume that spacing between sub-conductors is much larger than their radius, the maximum electric field intensity is experienced at

Y Z

X

Power Systems

The total reactance and total susceptance of a lossless overhead EHV line, operating at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity of wave propagation is km/s, then the approximate length of the line is (A) 122 km (C) 222 km (B) 172 km (D) 272 km

9.

EE-2008 10. An extra high voltage transmission line of length 300 km can be approximated by a lossless line having propagation constant =0.00127 radians per km. Then the percentage ratio of line length to wavelength will be given by (A) 24.24% (C) 19.05% (B) 12.12% (D) 6.06% A lossless transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be (A) 1835 MW (C) 2725 MW (B) 2280 MW (D) 3257 MW

11.

EE-2009 12. For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real power loss will be proportional to (A) V (C) 1/V2 (B) V2 (D) 1/V EE-2010 13. Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltages and are as indicated in the figure, and I > 0, then

W (A) Point X (B) Point Y

(C) Point Z (D) point W th

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GATE QUESTION BANK Power Flow AC System A

(A) (B) (C) (D) 14.

15.

Consider a three-core, three-phase, 50Hz, 11kV cable whose conductors are denoted as R, Y and B in the figure. The inter-phase capacitance ( ) between each pair of conductors is 0.2µF and the capacitance between each line conductor and the sheath is 0.4.µF. The per-phase charging current is

16. C

A

AC System B D

B Rectifier

, , , ,

Inverter

, , ,

Consider two buses connected by an impedance of (0+j5) . The bus1 voltage is 100 V, and bus2 voltage is 100 V. The real and reactive power supplied by bus 1, respectively, are (A) 1000 W, 268 VAr (B) 1000 W, 134 VAr (C) 276.9 W, 56.7 VAr (D) 276.9 W, 56.7 VAr

C1

Outer Sheath

(A) 2.0 A (B) 2.4 A

(C) 2.7 A (D) 3.5 A

Consider a three-phase, 50Hz, 11kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self-capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure. The voltage across the two insulators are

17.

C

(A) The magnitude of terminal voltage decreases, and the field current does not change. (B) The magnitude of terminal voltage increases, and the field current does not change. (C) The magnitude of terminal voltage increases, and the field current increases. (D) The magnitude of terminal voltage does not change, and the field current decreases.

Y C1

Receiving end

~

R

B

A 50Hz synchronous generator is initially connected to a long lossless transmission line which is open circuited at the receiving end. With the field voltage held constant, the generator is disconnected from the transmission line. Which of the following may be said about the steady state terminal voltage and field current of the generator? Long Transmission Line

Power Systems

5C

e

5C

e

Conductor

(A) (B) (C) (D)

e e e e

= 3.74 kV, e = 2.61 kV = 3.46 kV, e = 2.89 kV = 6.0 kV, e = 4.23 kV = 5.5 kV, e = 5.5 kV

EE-2011 18. A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options.

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GATE QUESTION BANK (A)

~ 132 kV, 300 km double circuit

(B)

~

Power Systems

All the line reactance’s are equal to j1 Bus 2 Bus 1(slack) j1

Load Center

~

Load center

~

P2=0.1 pu

132 kV, 300 km single circuit with 40% series capacitor compensation

(C)

~ 400 kV, 300 km single circuit

(D)

~ 400 kV, 300 km double circuit

Bus 3

P3=0.2 pu

Load center

EE-2013 19. A single–phase load is supplied by a single–phase voltage source. If the current flowing from the load to the source is 10 A and if the voltage at the load terminals is 1006 V, then the (A) Load absorbs real power and delivers reactive power (B) Load absorbs real power and absorbs reactive power (C) Load delivers real power and delivers reactive power (D) Load delivers real power and absorbs reactive power 20.

j1

j1

Load center

For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3 per unit. If a capacitor having reactance of j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is (A) 0.325 (B) 0.325 (C) 0.371 (D) 0.433 Statement for Linked Answer Questions 21 and 22 In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible.

21.

The voltage phase angles in rad at buses 2 and 3 are (A) θ = ,θ = (B) θ = , θ = (C) θ = , θ = (D) θ = ,θ =

22.

If the base impedance and the line–to line base voltage are 100 ohms and 100 Kv, respectively, then the real power in MW delivered by the generator connected at the slack bus is (A) 10 (C) 10 (B) 0 (D) 20

23.

(t) = cos A source t has an internal impedance of (4+j3) . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in should be (A) 3 (C) 5 (B) 4 (D) 7

EE – 2014 24. A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400V, 50 Hz balanced sources. Both voltage sources S and S are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure. S

m

~

m

z

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m

S

~ z

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GATE QUESTION BANK

The contributions of current supplied respectively, are (A) 75 A and 25 A (B) 50 A and 50 A 25.

26.

28.

A 183-bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain the load flow solution using Newton-Raphson method in polar coordinates, the minimum number of simultaneous equations to be solved is______________

29.

The complex power consumed by a constant-voltage load is given by ( j ), where, k k and k k A compensating shunt capacitor is chosen such that | | 0.25 kVAR, where Q is the net reactive power consumed by the capacitor-load combination. The reactive power (in kVAR) supplied by the capacitor is___________

30.

For a 400 km long transmission line, the series impedance is (0.0 + J0.5) /km and the shunt admittance is (0.0 + J5.0) mho/km. The magnitude of the series impedance (in ) of the equivalent circuit of the transmission line is____________

(C) 25 A and 75 A (D) 0 A and 100 A

Shunt reactors are sometimes used in high voltage transmission systems to (A) Limit the short circuit current through the line. (B) Compensate for the series reactance of the line under heavily loaded condition. (C) Limit over-voltages at the load side under lightly loaded condition. (D) Compensate for the voltage drop in the line under heavily loaded condition. In a long transmission line with r,l,g and c are the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is (A) rc = lg (C) rg = lc (B) r = √l⁄c

27.

S and S in 100 A at location P

Power Systems

(D) g = √c l

For a fully transposed transmission line (A) positive, negative and zero sequence impedances are equal (B) positive and negative sequence impedances are equal (C) zero and positive sequence impedances are equal (D) negative and zero sequence impedances are equal

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GATE QUESTION BANK

Power Systems

Answer Keys & Explanations 1.

2.

3.

[Ans. C] It is generally based wavelength of the line

5.

[Ans. A] a

~

[Ans. B] Rectifiers act as an inductor (for α < 900) Inverters act as a capacitor (for α < 900)

k

(

)

( (

)

∴ (%)

=0

ase impedance =

=

=(

(

)

= [Ans. C] +1 P1

=

=

pu

[Note: The Question in IIT Gate was wrong and later it is modified]

1 P2

6.

[Ans. A]

~

D2=3 m D1=2.5 m

Y

15kV

A

T1

T2 Where, P1and P2 form power line and T1 and T2 form telephone line Current in power circuit = I = 100 A Mutual inductance between power line and telephone line D = ln ( ) ⁄m D

=

=

)

=

= 16%

4.

k

ase impedance = = alues of load in ohm = = alue of load in p u ase impedance = = Base values in generator circuit ( ) = (k ) = k

)

At the time of N.L, ∴ ( )=

k

Base value on load circuit ( ) = and (k ) =

[Ans. C] % voltage regulation of the line =

load

line

ln (

B θ

Y 400kV

Y

220kV

Autotransforme

Taking V1 asrthe reference =  k θ = Phase difference and is So, =  k leads y So, =  k θ = lags y θ =θ θ =θ = = Phase difference between and is θ θ=θ θ = = So, option (A) is correct

)

= m Voltage induced in the telephone line =| |= = | | = m = km

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GATE QUESTION BANK

7.

8.

[Ans. C] Both regions are connected by HVDC link as well as AC line. So AC line is possible when both regions have same frequency. By changing fringe angle (α) of converter, we can change the power sharing, between the AC line and HVDC link.

θ= √

X

Point X

10.

2

=

= 4947.39 km

Ratio = = = 0.0606 % Ratio = 6.06 %

3

Point Y 1

1

km

[Ans. D] Wave length = λ, propagation constant = λ=

1

3

l l

l = length of line =

11. 2

l =

[Ans. B] Electric field intensity at various points are shown as follows.

1

Power Systems

[Ans. C] Surge impedance loading = SIL = 2280 MW = Surge impedance Surge Impedance loading = SIL = as voltage is constant S

2

Point Z

3

2

3

Point W

Suppose Csc is the series capacitance per unit length for series compensation. Therefore, series reactance will be,

It is clear from the above diagrams that minimum cancellation of vector occurs at the point Y. Hence maximum electric field intensity.

9.

j

[Ans. C] Velocity of wave propagation √(

Therefore, f

Total inductance of line =

S

=S

S

=

Inductance/km = Total susceptance of line = 1.2 pu =

=j

(

=j

(

) )

) =j ( here is known as degree of series compensation

(θ) = )( ) km km Let l is the length of the line Total reaction of line =0.045 pu =

j

fc

12.

Total capacitance of line = apacitance km =

√

(

=√ =

)

√

=

[Ans. C] S = Complex power S is defined as , S = VI S = If R is the resistance of transmission line

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GATE QUESTION BANK

Real power loss =

16.

=( ) =

Power Systems

[Ans. A]

S

0.4 F

Real power loss

13.

14.

Virtually connected to sheath

0.6 F

Since, S and R are fixed

0.6 F

0.6 F 0.4 F

[Ans. C] Because, power flow takes from AB to CD so , , and Since current flows from high voltage to low voltage.

0.4 F

Per charging current =j | | = =

[Ans.A] =  and =  as , current will flow from bus 1 to bus 2 = j 

17.

√

[Ans. B]

= =





I1

=  omplex power = S = ( j =  = j = and =

15.

j = 

)

5C

e

Line to line voltage = = k = Phase to ground voltage

[Ans. A]  As field voltage is held constant, so field current does not change.  When the generator is connected with open – circuit transmission line, line draws charging current

Therefore is higher than Eg i.e. Vt > Eg But when the generator is disconnected from the line, no charging current is delivered by the generator i.e. IC = 0. In this Vt = Eg So, terminal voltage decreases.

e

I2 I

C

r

5C

=

√ =

e

√

e =

k

√

k =

k

= (j ) e = (j ) e (j ) e = e Solving equation (i) and (ii), we get e = k and e = k

18.

th

(i) e

(ii)

[Ans. D]  For transmission of bulk power of very long distance high voltage (400 kV) is used.  To increase reliability, double circuit is used. th

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GATE QUESTION BANK

19.

21.

[Ans. B] Current from load to source is

[Ans. B] Bus-1 slack

~

Current from source to load = = =

j = =

u

u

x = [x = ] x = Here all the bus voltage are assumed at 1 P.u so, it is a DC load flow problem θ So, [ ] = [ ] [ ] θ and are power injection at bus 2 and 3 = = [load is ve] x

= [

x x

x x

=*

x ] +=*

θ So, [ ] = * θ

[Ans. D] Equivalent circuits

j j

+

+ +=*

*

+

[Ans. C] From previous solution we got. θ [ ]=* + [ ]=* + θ Now. So, = = = Now, Base VA=

j

~

Bus-3

22.

=

Bus-2

j

j

Current lags voltage so, it absorbs the real and reactive power. (or) Alternative Method: Current from source to load = = S= =( )( ) = j = j ∴ , Hence load absorbs real power and reactive power

20.

Power Systems

(

)

=

Base MVA=100 So, MW delivered by slack generator (bus – ) = =

=

23.

th

[Ans. C] (t) = cos t internal impedance = ( j ) For maximum power transfer, the load impedance magnitude should be same th

th


GATE QUESTION BANK

as source internal thevinen’s impedance magnitude Given the output should be only resistive in nature maximum power transfer will be at =| i| =√

24.

27.

[Ans. B] Positive sequence impedance = negative sequence impedance = x x Zero sequence impedance = x x x = self-reactance of each line x = mutual reactance of any line pair

28.

[Ans. 332] Number of equation to solved is n = = Where n number of PQ buses m number of PV buses

=

[Ans. D] m m

z

m

m

z

26.

m

2

From source 1, current I is supplied Let the resistance per km By KVL, 400 +I (400R) + (I 200)200R + (I 300)200R+ (I 500)200R +400=0 400+400IR+200IR 40000R+200IR 60000R+200IR 1000R+100=0 On solving I = 200A Which means at point P contribution of source 1 is 200 – 200A = 0A And contribution of source 2 to point P is rest 100A

25.

Power Systems

29.

[Ans. 0.75] Reactive power supplied = =Initial maximum reactance power drawn by load = Net reactive power consumed by capacitor load combination = =

30.

[Ans. *] Range 186 to 188 z= j km y= j km = km section series impedance =

[Ans. C] Under lightly loaded condition leakage capacitance will supply to the load end causing its voltage to rise. Hence shunt reactors are used to maintain the voltage at receiving by absorbing some reactive power.

= zl =

j

r = √zy = √ rl = (

sin hrl rl

j

j

)j√

series impedance =

(

) (

(√ )(√

(

))

)

= j ∴ Magnitude of series impedance =

[Ans. A] For distortion less line = rc = lg

th

th

th


GATE QUESTION BANK

Power Systems

Economics of Power Generation EE -2007 1. The incremental cost curves in Rs/MW hr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is: Incremental Cost Rs/MWhr 600 450

200 MW 450 MW GENERATOR A

P

EE-2008 3. A lossless power system has to serve a load of 250 MW. There are two generators ( and ) in the System with cost curves and respectively defined as follows: ( )= + 0.055 × ( )=3 + 0.03 × Where and are the MW injections from generator and respectively. Then the minimum cost dispatch will be (A) = 250 MW; = 0 MW (B) = 150 MW; = 100 MW (C) = 100 MW: = 150 MW (D) = 0 MW; = 250 MW

Incremental Cost Rs/MWhr

EE-2009 4. Three generators are feeding a load of 100MW. The details of the generators are

800 650

400 MW 150 MW GENERATOR B

P

(A) Generator A: 400 MW, Generator B: 300 MW (B) Generator A: 350 MW, Generator B: 350 MW (C) Generator A: 450 MW, Generator B: 250 MW (D) Generator A: 425 MW, Generator B: 275 MW 2.

An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz. If an additional load of 3.5MW is connected then the frequency will settle down to (A) 49.417 Hz (C) 50.083 Hz (B) 49.917 Hz (D) 50.583 Hz

Generator-1 Generator-2

Rating (MW) 100 100

Efficiency (%) 20 30

Regulation (p.u.) on 100 MVA base 0.02 0.04

Generator-3

100

40

0.03

In the event of increased load power demand, which of the following will happen? (A) All the generators will share equal power (B) Generator-3 will share more power compared to Generator-1 (C) Generator-1 will share more power compared to Generator-2 (D) Generator-2 will share more power compared to Generator-3 EE-2011 5. A load center of 120 MW derives power from two stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the figure below. The three generators and are of 100 MW capacity each and have identical fuel cost characteristics. The minimum loss th

th

th


GATE QUESTION BANK

generation schedule for supplying the 120 MW load is

~

25 km

120 MW

(A)

(B)

(C)

(D)

6.

= 80 MW + losses = 20 MW = 20 MW = 60 MW = 30 MW + losses = 30 MW = 40 MW = 40 MW = 40 MW + losses = 30 MW + losses = 45 MW = 45 MW

For enhancing the power transmission in a long EHV transmission line, the most preferred method is to connect a (A) series inductive compensator in the line (B) shunt inductive compensator at the receiving end (C) series capacitive compensator in the line (D) shunt capacitive compensator at the sending end

EE-2012 7. The figure shows a two-generator system supplying a load of PD = 40 MW, Connected at bus 2. Bus 1

~

Bus 2

p , where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is (A) (B) (C) (D)

~ ~

75 km

EE-2014 8. Three-phase to ground fault takes place at locations and in the system shown in the figure If the fault takes place at location , then the voltage and the current at bus A are and respectively. If the fault takes place at location , then the voltage and the current at bus A are and respectively. The correct statement about voltages and currents during faults at and is B

~ (A) (B) (C) (D)

~

leads and leads leads and lags lags and leads lags and lags

9.

In an unbalanced three phase system, phase current pu, negative sequence current pu, zero sequence current pu. The magnitude of phase current in pu is (A) 1.00 (C) 11.53 (B) 7.81 (D) 13.00

10.

A two bus power system shown in the figure supplies load of 1.0+j0.5 p.u.

~

The fuel cost of generators and ( ) = 10,000 Rs/MWh and ( ) =12,500 Rs/MWh And the loss in the line is

Power Systems

are:

th

th

th


GATE QUESTION BANK G1

~

j

13.

The horizontally placed conductors of a single phase line operating at 50 Hz are having outside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. The permittivity of free space is /m. The capacitance to ground per kilometer of each line is (A) (C) (B) (D)

14.

A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The positive, negative, and the zero sequence reactances of the generator are 0.2 pu and 0.05 pu, respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at the terminal of the unloaded generator, the fault current in amperes immediately after the fault is______________

j

j

The values of respectively are (A) 0.95 and 6.00 (B) 1.05 and 5.44 11.

12.

in p.u. and (C) 1.1 and 6.00 (D) 1.1 and 27.12

A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100. MVA and 437.38 A respectively. The line-to-line load voltage in kV is____________

Power Systems

A synchronous generator is connected to an infinite bus with excitation voltage = 1.3 pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is ___________

th

th

th


GATE QUESTION BANK

Power Systems


2.

[Ans. C] Maximum incremental cost in Rs/MWhr for Generator A =600 (at 450 MW) Minimum incremental cost in Rs/MWhr for Generator B =650(at 150 MW) As maximum value of incremental cost of A is less than minimum value of B. ∴ enerator ‘A’ will operate at it maximum (o/p) 450 MW and B at (700 450) =250MW

∝ Power shared by Generator 1> Generator 2> Generator 3 5.

[Ans. A]

~

[Ans. C] +

= 250MW

6.

[Ans. C] To enhance the power transmission in a long EHV transmission line, a series capacitor is used

→ (i)

= 1 + 0.11

Power flow in line ∝

= 3 + 0.06

*

+

For k% series capacitor compensation

For maximum cost of generation, = 1 + 0.11 = 3 + 0.06 → ii) Solving equation (i) and (ii), = 100 MW = 150 MW 4.

~ ~

Generator is near to load and , are away from the load. To reduce losses, more power is contributed from Hence = 80 + losses, = 20 MW, = 20 MW

3.5 = 1.16%

∴ ∆f = 0.583 ∴ f = 50 0.583 = 49.417 Hz

3R

R

120 MW

[Ans. A] Here, change in (freq.) w.r.t. power ( )

3.

Power shared by a generator

k

( 7.

)

[Ans. A] From coordination equation

[Ans. C] Let x1, x2 and x3 are reactances of generator -1, generator -2 and generator -3 respectively. Neglecting armature resistance of all the three generators. VR1=0.02p.u. VR2 =0.04 p.u and VR3=0.03 p.u. VR1 < VR3
Given an

p th

th

th


GATE QUESTION BANK

As the base value is 100 MVA

10.

A

( )

Power Systems

[Ans. B] j j j

p

j

j A

(

j j

co

j

in

co co

A 8.

)

j

in in

[Ans. C] For fault at Equivalent circuit A

tan

in 11. X X → Line impedance

[Ans. *] Range 117 to 120 co p √ power p √

Base

p √

|X |

X lea or fa lt at f

olt 12. lag 9.

[Ans. C] [ ]

[

[Ans. *] (Range 0.1 to 0.2) in X in in

][

]

j j

(

j

)

j co co

j in j

in

co ol ing thi | | th

th

th


GATE QUESTION BANK

13.

Power Systems

[Ans. B] ln (

R

)

R

ln (

)

km 14.

[Ans. *] Range 15300 to 15500 Reference

E

p j

j

j

j p a e p

√

√ a e

√

A

th

th

th


GATE QUESTION BANK

Power Systems

Symmetrical Components & Faults Calculations EE-2006 1. Three identical star connected resistors of 1.0pu are connected to an unbalanced 3 phase supply. The load neutral is isolated. The symmetrical components of the line voltages in P.u are: =X , =X . If all the P.u calculations are with the respective base values, the phase to neutral sequence voltages are (A) =X ( + ), =Y ( ) (B) =X ( ), =Y ( + ) (C)

= =

(D)

=

√ √

√

=

√

X (

),

Y (

)

X (

),

Y (

)

Common data Question For 2 and 3. A generator feeds power to an infinite bus through a double circuit transmission line. A 3- phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0pu power prior of the fault with rotor power angle of . The system frequency is 50 Hz. 2.

3.

If the initial accelerating power is X pu, the initial acceleration in elec. deg/sec2, and the inertia constant in MJ – sec /elec. deg respectively will be (A) 31.4 X,18 (C) X/1800,0.056 (B) 1800 X,0.056 (D) X/31.4,18 Common Data Question for 4 and 5 For a power system the admittance and impedance matrices for the fault studies are as follows. j j j j j =[ j ] j j j j j j j j = [j ] j j j The pre-fault voltages are 1.0 p.u at all the buses. The system was unloaded prior to the fault. A solid 3 phase fault takes place at bus 2.

4.

The post fault voltages at buses 1 and 3 in per unit respectively are (A) 1.24, 0.63 (C) 0.33, 0.67 (B) 0.31, 0.76 (D) 0.67, 0.33

5.

The per unit fault feeds from generators connected to buses 1 and 2 respectively are (A) 1.20, 2.51 (C) 1.66, 2.50 (B) 1.55, 2.61 (D) 5.00, 2.50

6.

The Gauss Seidel load flow method has following disadvantages. Tick the incorrect statement (A) unreliable convergence (B) Slow convergence (C) Choice of slack bus affects convergence (D) A good initial guess for voltages is essential for convergence

The initial accelerating power (in pu) will be (A) 1.0 (C) 0.56 (B) 0.6 (D) 0.4

th

th

th


GATE QUESTION BANK

EE-2007 7. A three phase balanced star connected voltage source with frequency ω rad/s is connected to a star connected balanced load which is purely inductive. The instantaneous line currents and phase to neutral voltages are denoted by (i i i ) and ( ) respectively and their rms values are denoted by V and I. [ ] If √ √

9.

A 230 V (phase), 50 Hz, three – phase, 4 – wire systems has a phase sequence ABC. A unity power-factor load of 4kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and pure capacitor in the other two phases. The value of inductor and capacitor is (A) 72.95 mH in phase C and 139.02 µF in phase B (B) 72.95 mH in phase B and 139.02 µF in phase C (C) 42.12 mH in phase C and 240.79 µF in phase B (D) 42.12 mH in phase B and 240.79 µF in phase C

i [i ] i

√ √

[ √ ] √ Then the magnitude of R is (A) 3 VI (C) 0.7 VI (B) VI (D) 0 8.

Suppose we define a sequence transformation between “a-b-c” and “p-n-o” variables as follows: f f [f ] k [ ] [f ] where e f f and k is a constant. Now, if it is given that : i [ ]=[ ] [i ] and i i [ ] [i ] then, i (A) Z = [

*

X

+[

Y F

An 'a' phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase 'a', are given by Va Volts and Ia Amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by

]

(C) Z = 3k2 [ (D)

EE-2008 10. A two machine power system is shown below. Transmission line XY has positive sequence impedance of Z1 Ω and zero sequence impedance of Z0 Ω

]

(B) Z = [

Power Systems

]

(A)

Ω

(C)

(B)

Ω

(D)

Ω Ω

]

th

th

th


GATE QUESTION BANK

11.

A 3 - phase transmission line is shown in the figure

Power Systems

Common Data Question for 13, 14 and 15 Consider a power system Y

X

~

I

I

Voltage drop across the transmission line is given by the following equation I [ ] [ ] [I ] I Shunt capacitance of the line can be neglected. If the line positive sequence impedance of Ω and zero sequence impedance of Ω then the values of s and Zm will be (A) Zs Ω; m Ω (B) Zs Ω; m Ω (C) Zs Ω; m Ω (D) Zs Ω; m Ω Single line diagram of a 4-bus single source distribution system is shown below. Branches e1, e2, e3, and e4 have equal impedances. The load current values indicated in the figure are in per unit Distribution Company’s policy requires radial system operation with minimum loss. This can be achieved by opening of the branch

~ e

1+j0

13.

The instant (t0) of the fault will be (A) 4.682 ms (C) 14.667 ms (B) 9.667 ms (D) 19.667 ms

14.

The rms value of the ac component of fault current (Ix) will be (A) 3.59 kA (C) 7.18 kA (B) 5.07 kA (D) 10.15 kA

15.

Instead of the three phase fault, if single line to ground fault occurs on phase 'a' at point ‘F’ with zero fault impedance then the rms value of ac component of fault current(Ix)for phase 'a' will be (A) 4.97pu (C) 14.93pu (B) 7.0pu (D) 29.85pu

e

e

e 5+j0 2+j0

(A) e (B) e

~ I

Given that: = =1.0 + j0.0 P.u; The positive sequence impedance are = =0.001 + j0.01 P.u and = 0.006 + j0.06 P.u 3 phase Base MVA = 100 Voltage base = 400 kV (Line to Line) Nominal system frequency = 50 Hz The reference oltage for Phase ‘a’ is defined as V(t)=Vm cos(t). A symmetrical three phase fault occurs at centre of the line i e point ‘F’ at time t0. The positive sequence impedance from source to point ‘F’ equals (0.004 + j0.04)p.u. The waveform corresponding to phase 'a' fault current from bus X reveals that decaying dc offset current is negative and in magnitude at its maximum initial value. Assume that the negative sequence impedances are equal to positive sequence impedances, and the zero sequence impedances are three times positive sequence impedances.

I

12.

F

I

(C) e (D) e

th

th

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GATE QUESTION BANK

EE-2009 16. For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt elements are.

Power Systems

(A)

R G

(B) j[

17.

G

(C) 1 and 2 (D) 1, 2 and 4

(C)

Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists. A. B. C.

D.

List I Improve power factor reduce the current ripples increase the power flow in line reduce the Ferranti effect

Codes: A (A) 2 (B) 2 (C) 4 (D) 4

B 3 4 3 1

C 4 3 1 3

1.

List II shunt reactor

2.

shunt capacitor

3.

series capacitor

4.

series reactor

r G

R

r G

19.

D 1 1 2 2

For the power system shown in the figure below, the specifications of the components are the following: G : 25 kV, 100 MVA, X = 9% G : 25 kV, 100 MVA, X = 9% T : 25 kV/220 kV, 90 MVA, X = 12% T : 220 kV/25 kV, 90 MVA, X = 12% Line 1: 220 kV, X = 150 ohms T

T Line 1

G

Bus 1

~

~

Bus 2

G

Choose 25kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is

r

(A)

b

j 0.27

j 0.42

j 0.27

j 0.18

Y B

R

(D)

EE-2010 18. The zero-sequence circuit of the three phase transformer shown in the figure is R

r

R

]

(A) 3 and 4 (B) 2 and 3

r

G

j 0.18

~

~

G

y

(B)

j 0.27

j 0.62

j 0.27

j 0.18 G

th

th

j 0.18

~

~ th

G


GATE QUESTION BANK (C)

j 0.42

j 0.27

(C)

j 0.27

j 0.21

1

j0.10

j 0.21

j2.25

3

j2.25

2

~

j0.10

~

~

~

G

Power Systems

G (D)

(D)

j 0.42

j 0. 3

j 0.3

~

~

~ G115 kV

j0.10

~

22.

A three-bus network is shown in the figure below indicating the p.u. impedance of each element. 1 2 3

15 kV

j0.2

G2

j0.08

j0.1

G1 = 250 MVA , 15 kV , positive sequence reactance X = 25% on its own base G2 = 100 MVA , 15 kV , positive sequence reactance X = 10% on its own base and = 10 km , positive sequence reactance X = 0.225 /km

j0.1

The bus admittance matrix, Y-bus, of the network is (A) j[

]

(B) j[

]

(C) j[ 20.

2

In the above system, the three-phase fault MVA at the bus 3 is (A) 82.55 MVA (C) 170.91 MVA (B) 85.11 MVA (D) 181.82 MVA

~

10 km

10 km

j2.25

21.

2

3

3

G

EE-2011 Statement for Linked Answer Question 20 & 21 Two generator units G1 and G2 are connected by 15 kV line with a bus at the mid-point as shown below, 1

j2.25

~

j 0.21

j 0.21 G

1

j0.25

For the above system, the positive sequence diagram with the p.u values on the 100 MVA common base is

]

(D) j[

]

(A) 1

j0.10

j1.0

3

j1.0

2

~

j0.10

23.

~

(B) 1

j0.25

~

j1.0

3

j1.0

2

j0.10

A negative sequence relay is commonly used to protect (A) an alternator (B) a transformer (C) a transmission line (D) a bus bar

~

th

th

th


GATE QUESTION BANK

EE-2012 24. The bus admittance matrix of a three-bus three-line system is j[

EE-2014 27. A 2-bus system and corresponding zero sequence network are shown in the figure.

]

us

26.

~

b

and and C

and

D

and

Bus 2

pu

~

j

pu

(A) 0.2 pu (B) 0.268 pu

us

The transformers T and T are connected as

For the system shown below, and are complex power demands at bus 1 and bus 2 respectively. If |V2| = 1 pu, the VAR rating of the capacitor ( ) connected at bus 2 is Bus 1

T

a

The sequence components of the fault current are as follows: I = j 1.5 pu, I = j 0.5 pu, I zero = j1 pu. The type of fault in the system is (A) LG (C) LLG (B) LL (D) LLLG

v

T

~

If each transmission line between the two buses is represented by an equivalent -network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is (A) 4 (C) 1 (B) 2 (D) 0 25.

Power Systems

pu

pu

(C) 0.312 pu (D) 0.4 pu

th

th

th


GATE QUESTION BANK

Power Systems


[Ans. C] In positive sequence. Phase sequence is abc Assuming and

=1.1 p.u. Voltage of the bus = | | pu Equivalent transfer admittance = 0.8. Transfer reactance = x P

= Electrical power delivered by the generator

during fault = P

√ x √ x √ x

|

||

sin

sin

pu

Mechanical input do not change during fault So, P pu Accelerating power = P P P P pu

√ o if

=

|

3.

[Ans. B] Inertial constant (in MJ –s/elec deg) GH f Where, G = Machine rating = 10 MVA H = Inertial constant in MJ/MVA = 5 MJ/MVA GH f H f Accelerating in elec deg/ sec P P ccelerating power = x P x x

4.

[Ans. D]

√ In negative sequence, Phase sequence is acb Assuming and

√ √ So, if √ √ √

2.

[Ans. C] Before fault Electrical power delivered by the generator P pu Mechanical input to the generator = P At steady state, P P pu During fault Initial rotor angle = Internal voltage of the generator =|E |

Pre fault

[

]

[

]

Fault current =I Solid 3-∅ fault occurs at bus

r

and

j

th

th

th


GATE QUESTION BANK

I

6.

[Ans. A] 1. The time taken to perform one iteration of the computation is relatively smaller in case of G-S method but the number of iterations required by G-S method for a particular system and they increase with the increase in the size of the system. Which results in slow convergence. 2. The convergence characteristic of G-S method is sometimes very seriously affected by the selection of a slack bus and the selection of a particular bus may result in poor convergence.

7.

[Ans. A]

j j pu Post fault voltage at bus I is given by I Post fault voltage at bus I=1 I j j pu Post fault voltage at bus I = 3 I j j pu 5.

[Ans. C] Yii = sum of the admittance directly connected to ith bus y y y y excluded y o y y y y y y y j j j y j pu Similarly, y y j j j y j pu y

I

y E

j j

Power Systems

√

]

R=[

√

[ R =*

√

I

I

√

√

√

I

I [I ] I

√

]

√

I

I

√

I +

Magnitude of R is determined of matrix R R = 3VI 8.

[Ans. B] Here, [

]

I ] [I ] I

[

j Where, [

j

I ] = [Z] [I ] I I ] [I ] I

I [I ] = K [ I

[Current fed by generator i during fault] E E [The system was unloaded] Current fed by generator 1

I [I ] I Similarly, = K[A]

I pu Current fed by generator 2,

= K[A] [

] [I

= K[A] [

]

I pu th

th

th

[

] ]

I


GATE QUESTION BANK

Comparing (1) & (2) ∴

][

[

= [ =[

9.

10.

[Ans. D] As the ground distance relay is located at terminal X Impedance measured by the relay

11.

[Ans. B] Let & be the self and mutual inductance of a line Positive sequence impedance = Ω i Zero sequence impedance = + Ω ii Solving eq (i) and eq (ii) ∴ Ω Ω

12.

[Ans. D] Let impedance of each branch be R on removing (e ) Losses = Similarly on removing e , Losses = on removing e , Losses = on removing e , Losses = o on removing bus ‘e ’ losses are minimum, so correct option is (d)

13.

[Ans. A] Voltage, V = cos ωt Current after fault

]

][

][

Power Systems

]

]

[Ans. B] I

I I

Taking

as the reference

P cos

I

 If an pure inductor is present in phase B, then I lags by  If a pure capacitor is present in phase C, then I leads by If current through neutral is to be zero I sin I sin I I I I I cos I cos I cos |I |

|I |

|I|

i t

√ I |I |

√ cos ωt | |

e

at t = t ; i = 0

√

e

√ cos ωt | |

⁄

j

√ |I |

|I |

|

|

| | The maximum value will be at ωt

ω

|

mH | ωC

t

ω

C C

f m sec

F th

th

th


GATE QUESTION BANK

14.

Power Systems

[Ans. A]

j .003 + j 0.03

.003 + j 0.03

.001 + j 0.01

.001 + j 0.01

eq

~

j j j From eq (i)

~

= =

j

j

j

j

j

From eq (ii)

I =

= =

eq = j || j eq = 0.002 + j 0.02 p.u E = = 1.0 p.u [ pre fault voltage not specified] ∴I =

= 49.75

I

=

I

=I

j

j

j

j

j

j

From eq (iii) pu

= =

j

= =

j

From eq (iv)

= 144.34

√

j j

I

j

j

= 7.18KA I = 15.

= 3.59

[Ans. C] I = Given that I =

=

and

=

I = 29.85 I = 16.

=

p.u 17.

= 14.93

[Ans. C] If directly we can add all the elements of each row. If the sum is not zero, then that bus contains shunt elements For ith bus add all the element of ith row. If sum = 0 ith bus contain shunt elements [

]

here j . . . .(i) j ii j . . (iii) j iv

[Ans. B]  Shut capacitor are used to provide part of the reactive ’s required by the load to keep the voltage within desirable limits and to improve factor.  Series reactor reduces current ripple.  Shunt reactors are used across capacitive loads or lightly loaded lines to absorb some of the leading ’s to control the voltage across the load to within certain desirable limit.  Series capacitor campensation reduces the series impedance of the line. Power flow inline increases, as

power flow in line decreases.

j j th

th

th


GATE QUESTION BANK

18.

[Ans. C]

G

19.

23.

[Ans. A] Negative sequence load used to protect the alternator against unbalanced load

24.

[Ans. B]

r

R

Power Systems

[Ans. B]

y

j

j[

]

j[ 

but, So, 20.

]

for Bus admittance matrix,

= j0.62 pu.

[Ans. A] J1.0

so shunt susceptance of the line connecting bus1 and 2 is+j2

J1.0

1

2 -10

+ j2

+j2 -10

3

N

N = 0.25 ×

25.

[Ans. C] I = +j 1.5 P.u I = j 0.5 P.u I = j 1.0 P.u Since, two sequence current is present, the type, of fault is not L-L fault In L-G fault: all sequence networks will be connected in series. So, the current is same for all I I &I In for the case of LLLG fault, as it is balanced fault, only two sequence current will be present in the system. So, from the above, the type of fault is LLG

26.

[Ans. B]

( ) = 0.1

= 0.1 ×

21.

( ) = 0.1

= 0.225 × 10 (

) = j1.0

= 0.225 × 10 (

) = j1.0

[Ans. D] C

22.

[Ans. B] 1

2

j

j

j0.1

j j5 0

3

j0.08

j0.1

j

=

j

V1 =

j0.2

y

+j 2

j j

j5 j

~

j

= 1p.u.

j

1-0

z= j0.5p.u.

0 j j

V2=

I12

Line is lossless

= 1p.u.

= 1+1 = 2p.u. Power transfer from bus-1 to bus-2 is th

th

=

th

+


GATE QUESTION BANK

Power Systems

l p.u. |

∴

||

|

sin (

; sin sin ∴

2

=

sin ;

}; ; V2 = 1

I12 =

= 1-j0.288

Current =l ; Current in =l [1 j0.268]= 0.268 VAR rating of capacitor = |V2||lQ|sin(|V2||I2|) = 1 0.268 sin(+90) = 0.268 27.

[Ans. B] For

New sequence network For

th

th

th


GATE QUESTION BANK

Power Systems

Power System Stability EE-2006 1. A 400 V, 50Hz, three phase balanced source supplies power to a star connected load whose rating is 12√3 k VA, 0.8 pf (lag). The rating (in kVAR) of the delta connected (capacitive) reactive power bank necessary to bring the pf to unity is (A) 28.78 (C) 16.60 (B) 21.60 (D) 12.47 EE-2007 2. Consider the two power systems shown in figure A below, which are initially not interconnected, and are operating in steady state at the same frequency. Separate load flow solutions are computed individually for the two systems, corresponding to this scenario. The bus voltage phasors so obtained are indicated on figure A. These two isolated systems are now interconnected by a short transmission line as shown in figure B, and it is found that P1 = P2 = Q1 = Q2 = 0; 1 02 0

~ ~

X

1 02 10

~ ~

1 0 20

Fig (A) P1,Q1

X

10

1 02 0

5

x

1.0 0

x

10 One line trips 130

t

(A) 0.87 (B) 0.74 4.

(C) 0.67 (D) 0.54

The figure below shows a three phase selfcommutated voltage source converter connected to a power system. The converter’s dc bus capacitor is marked as C in the figure. The circuit is initially operating in steady state with δ = 0 and the capacitor dc voltage is equal to Vdc0. You may neglect all losses and harmonics. What action should be taken to increase the capacitor dc voltage slowly to a new steady state value?

Y 1 02 15

~ ~ Y

The bus voltage phase angular difference between generator bus X and generator bus Y after the interconnection is (A) 100 (C) 300 0 (B) 25 (D) 300 3.

0.1 pu

~

~ ~

P2,Q2

Fig (B)

it is constant at 1.0 pu. Due to some previous disturbance, the rotor angle (δ) is undergoing an undamped oscillation, with the maximum value of δ(t) = 1300. One of the parallel lines trips due to relay malopeation at an instant when δ(t) = 1300 as shown in the figure. The maximum value of the power unit line reactance, x, such that the system does not lose synchronism subsequent to this tripping is

Three Phase Voltage Source converter

C

δ

0

(A) Make δ positive and maintain it at a positive value (B) make δ positive and return it to its original value

Consider a synchronous generator connected to an infinite bus by two identical parallel transmission lines. The transient reactance of the generator is 0.1 pu and the mechanical power input to th

th

th


GATE QUESTION BANK

(C) make δ negative and maintain it at a negative value (D) make δ negative and return it to its original value

peed orque M

EE-2009 5. A 500 MW, 21kV, 50 Hz, 3-phase, 2- pole synchronous generator having a rated p.f = 0.9, has a moment of inertia of 27.5 10 kg-m2. The inertia constant (H) will be (A) 2.44 s (C) 4.88 s (B) 2.71 s (D) 5.42 s EE-2012 6. A cylindrical rotor generator delivers 0.5 pu power in the steady state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1pu. The inertia constant of the generator is 5MW-S/MVA generator reactance is 1 pu, the critical clearing angle, in degrees, for a three –phase dead short circuit fault at the generator terminal is (A) 53.5 (C) 70.8 (B) 60.2 (D) 79.6

(a) M

peed

orque (b)

The stable operating points are (A) P and R (C) Q and R (B) P and S (D) Q and S 9.

The figure shows the single line diagram of a single machine infinite bus system.

Infinte bus The inertia constant of the synchronous generator = 5M -s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle δ equal to 30 . A three phase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is sin δ If the values of δ and dδ⁄dt at the instant of fault clearing are 45 and 3.762 radian/s respectively, then (in pu) is____________

EE-2014 7. The undesirable property of an electrical insulating material is (A) high dielectric strength (B) high relative permittivity (C) high thermal conductivity (D) high insulation resistivity 8.

Power Systems

The torque-speed characteristics of motor ( ) and load ( ) for two cases are shown in the figures (a) and (b). The load torque is equal to motor torque at points P, Q, R and S

th

th

th


GATE QUESTION BANK

Power Systems


2.

3.

[Ans. D] = 12 √3 0 8 = 16.627 kW = 12 √3 0 6 =12.47 kW cos = 0 8 sin = 0 6 For unity pf, the total reactive power is zero Assuming kVAR rating of capacitor bank = =0 = = 12 47 k So, capacitor banks supplies 12.47 kVAR reactive power to the load. [Ans. A] As = = 0 and = =0 There is no energy transfer, so both bus bar voltage and its phase angle should be same, so we can consider both bus bar phase angle as zero. Phase angle of y y = 20 hase angle of x = x = 30 (∴ hase angle difference = x y = 30 20 = 10 [Ans. C] Here the alternator is undergoing under damped oscillations with δ max = 1300, at this any fault occurs even then also to maintain stability ‘ ’ value must not go beyond 1300 for a given mechanical input

= 10 0

E = 1.0 PS = 1.0 P.u Xeq: equivalent reactance when online is removed due to normal operation of relay Xeq = 0.1 + x ∴

(

sin 130 = 1 0

)

= 0 67 u [∵

S

= 1.0 P.u]

4.

[Ans. D] To increase the capacitor d.c. voltage slowly to a new slowly to a new steady state value, make ‘δ’ negative and return it to its original value.

5.

[Ans. A] G=

=

MW = 555.56 MVA

KE Stored =

M

(

=

27 5

) 10

(

)

= 1357.07 MJ H= 6.

=

= 2.44 MJ/MVA

[Ans. D]

~

δ

= 05 =

Infinite bus

| || | (

)

~

0

sin δ

15 1 sin δ (1 0 5)

δ = sin (0 5) = 30 = = 130

∴

=

,

sin δ =

6 Critical clearing angle δ = cos [( 2δ ) sin δ

δ

cos δ ]

S

th

th

th


GATE QUESTION BANK

= cos = cos

1 √3 + 3 3 2 (0 18) = 79 56 *(

)

7.

[Ans. B]

8.

[Ans. B] At P point, speed change by s, will increase - increase Hence speed will increases by s, hence move back to previous point. Hence stable point At s point, speed change by s, increases, speed will decrease by s. Hence move back to pervious point. Hence stable point d d d d For stability

9.

f

Power Systems

d δ (δ 30) = dt 180

(cos δ cos 30) dδ at δ = 45 , = 3 762 dt 5 3 762 3 14 50 (δ 30) √3 = (cos δ ) 180 2 5 3 762 3 14 50 (45 30) √3 = (cos 45 ) 180 2 0 12 = 0 88

1 ( √2

√3 ) 2

= 0 23

[Ans. *] Range 0.22 to 0.24 d δ = (pu) f dt = 1(pu) efore fault condition sin δ = (at equilibrium) ( ) sin δ = 1 sin 30 = 1 = 2(pu) sin δ

sin δ

0

30 45

During fault condition d δ = f dt d δ =1 sin δ f dt d δ = ∫ (1 sin δ)dt f dt

th

th

th


GATE QUESTION BANK

Power Systems

Protection & Circuit Breakers EE-2006 1. Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching (A) Vacuum (C) SF6 (B) air blast (D) Oil 2.

EE-2008 4. A lossless single machine infinite bus power system is shown below 1.0 δ pu

1.0 pu

The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power. Then the critical clearing time of the circuit breaker will (A) reduce to 0.14 s (B) reduce but will be more than 0.14 s (C) remain constant 0.28 s (D) increase beyond 0.28 s 5.

Voltage phasors at the two terminals of a transmission line of length 70 km have a magnitude of l.0 per unit but are 180 degrees out of phase. Assuming that the maximum load current in the line is th of minimum 3-phase fault current. Which one of the following transmission line protection schemes will NOT pick up for this condition? (A) Distance protection using mho relays with zone-1 set to 80% of the line impedance (B) Directional over current protection set to pick up at 1.25 times the maximum load current (C) Pilot relaying system with directional comparison scheme (D) Pilot relaying system with segregated phase comparison scheme

3

1

Transmission Line

2

Bus C Stuck breaker

4 Bus A

6

F 5 Bus B

(A) 1, 2, 6, 7, 3, 5 (B) 1, 2, 5, 6, 7, 3

Transmission Line

pu

~

In a biased differential relay, the bias is defined as a ratio of (A) number of turns of restraining and operating coil (B) operating coil current and restraining coil current (C) fault current and operating coil current (D) fault current and restraining coil current

EE-2007 3. Consider the protection system shown in the figure below. The circuit breakers numbered from 1 to 7 are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker 4 fails to operate (“stuck breaker”). If the relays are coordinated correctly, a valid sequence of circuit breaker operation is

.

7

(C) 5, 6, 7, 3, 1, 2 (D) 5, 1, 2, 3, 6, 7

th

th

th


GATE QUESTION BANK

EE-2009 6. Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists. List I List II a. Short Line 1. Ohm Relay b. Medium Line 2. Reactance Relay c. Long Line 3. Mho Relay (A) a 2, b 1, c 3 (B) a 3, b 2, c 1

9.

Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coil is CT ratio 400/5

(C) a 1, b 2, c 3 (D) a 1, b 3, c 2

Reactor

(A) 1pu (B) 1pu 8.

CT ratio 400/5

(220+j0)A

EE-2010 7. Consider a step voltage wave of magnitude 1pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is A

Power Systems

(250+j0)A

Operating coil

(A) 0.17875 (B) 0.2A

(C) 0.375A (D) 60kA

EE-2014 10. The overcurrent relays for the line protection and loads connected at the buses are shown in the figure.

(C) 2pu (D) 3pu

~ 100A

A three-phase, 33kV oil circuit breaker is rated 1200A, 2000MVA, 3s. The symmetrical breaking current is (A) 1200 A (C) 35 kA (B) 3600 A (D) 104.8 kA

The relays are IDMT in nature having the characteristic . ime ultiplier etting t ( lug setting ultiplier) . The maximum and minimum fault currents at bus B are 2000 A and 500 A respectively. Assuming the time multiplier setting and plug setting for relay to be 0.1 and 5A respectively, the operating time of (in seconds) is_________

th

th

th


GATE QUESTION BANK

Power Systems


2.

[Ans. A] Vacuum circuit breakers are specially used for low cost switch having low fault interrupting capacity, but capable of large number of load switching operations without maintenance.

4.

[Ans. D] 0.5 P.u

~ The critical clearing time can be evaluated for a given critical clearing angle for the special case where electrical output is zero and mechanical input is constant so that the accelerating power is constant during fault. Note:- [If the accelerating power is not constant then we cannot evaluate the critical clearing time ] d dt [∴ electrical power delivered ( e) = 0 during fault] d d .t dt dt t

I2

Electrical equipment

i2

i1

Relay coil

i0=|i1|-|i2|

Restraining coil

Operation of force relay coil = (|i | |i |) Opposing force produced by restrain coil =n (

| | | |

) From initial conditions at t = 0,

At balance (or) boundary condition (|i | =

|i |) = n ( | | | | | | | |

(

)

| | | |

( )

∴

)

t

= bias

where ritical clearing angle t ritical clearing time

Where n & n are the number of turns on operating relay coil and restrained coil respectively 3.

.

0.5 P.u

[Ans. B] I1

.

~

t

[Ans. C] If the relays are co – ordinated correctly, due to fault in a particular section, relay in that section must operate first, then relays in near by section if first fails, it is back up protection. Thus, the sequence is [5, 6, 7, 3, 1, 2].

√

(

)

Initially when one machine is delivering 1.0 P.u electrical output . .u When two identical generators are delivering a total electrical power output of 1.0 p.u., . p. u. t

th

.

th

√

(

)

th


GATE QUESTION BANK

√

t

(

)

(

)

√

The instant the travelling wave reaches the reactor (t)| p. u 8.

[Ans.C]

6.

7.

[Ans. A] Distance protection using mho relays with zone-1 set to 80% of the line impedance will not provide protection.

9.

)

k

√

CT ratio 400/5

250 A

250 A

CT ratio 400/5

I2

I1

I0C Operating coil

I

(

)

.

I

(

)

.

Current through operating coil = I I I . . . 10.

ZC

(in

)

[Ans. C]

[Ans. B] Let surge impedance of the line = Zc E=Step voltage wave of magnitude 1 p.u. s

) (

√

=

[Ans. A] For a short line of transmission, reactance relay is used For a medium line of transmission impedance relay is used For a long line of transmission, MHO relay is used.

(s)

(

I

∴ critical clearing time of breaker will increase beyond 0.28 sec. 5.

Power Systems

[Ans. *] Range 0.21 to 0.23

~ I(s)

E(s)

Ls

V(s)

.

t I(s) (s) (s)

I I

(s) s oltage across the reactor (s) I(s). s s s s

(s)

(t)| .e at t=0

(

)

(

)

(

)

.

aximum load current at or a setting current of plug setting is ax fault current ratio current setting

s

(

or elay [ . . . ] . .

. .

s) ⁄

. . . ( . . ) .

t

. ( .

th

th

th

. ) .

.

sec


GATE QUESTION BANK

Power Systems

Generating Stations EE-2009 1. Out of the following plant categories (i) Nuclear (iii) Pump Storage (ii) Run-of-river (iv) Diesel The base load power plants are (A) (i) and (ii) (B) (ii) and (iii) (C) (i), (ii) and (iv) (D) (i), (iii) & (iv)

constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio of load shared by plants and is (A) 1:4 (C) 3:2 (B) 2:3 (D) 4:1

3.

There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1 Hz/MW. Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is ______

EE-2014 2. The fuel cost functions of two power plants are

Where, and are the generated powers of two plants, and A and B are the


[Ans. C]

2.

[Ans. D]

From

3.

[Ans. *] Range 49.9 to 50.1 ( )

,

th

th

th


GATE QUESTION BANK

Power Electronics

Basics of Power Semiconductor Devices EE-2006 1. A voltage commutation circuit is shown in figure. If the turn off time of the SCRs is 50 sec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation?

200V dc. The forward drops of all transistors/diodes and gate-cathode junction during ON state is 1.0 V. The resistance R should be

3.

1.0Ω

50Ω C

+

100V

200V Th1

Th2

(A) 2.88 F (B) 1.44 2.

L

PT

+10V 50Ω

SCR

R

(C) 0.91 (D) 0.72

An SCR having a turn ON time of 5 sec, latching current of 50 mA and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be

20Ω

5kΩ

100V 0.5H

(A) 251 sec (B) 150 sec

(C) 100 sec (D) 5 sec

EE-2007 Common Data for Question 3 & 4 A 1: 1 Pulse transformer (PT) is used to trigger the SCR in the adjacent figure. The SCR is rated at 1.5kV, 250A with IL=250 mA, IH =150 mA, and =150 mA, =100 mA. The SCR is connected to an inductive load, where L=150 mH in series with a small resistance and the supply voltage is

(A) 4.7 k Ω (B) 470 Ω 4.

(C) 47 Ω (D) 4.7Ω

The minimum approximate volt-second rating of the pulse transformer suitable for triggering the SCR should be: (volt-second rating is the maximum of product of the voltage and the width of the pulse that may be applied) (A) 2000 V-s (C) 20 V-s (B) 200 V-s (D) 2.0 V-s

EE – 2008 5. The truth table of a monoshot shown in the figures is given in the table below: Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure. The pulse widths of the two monoshot outputs, Q and Q are T and T respectively. R X

Y

Q

̅ Q X Y

0 1

th

th

C

th

T

0.7R

T

07R

Q ̅ Q


GATE QUESTION BANK R

C Q

X

T

0.7R

+5V

(B)

C Q

T

0.7R ̅ Q

̅ Q

1.5 current

R

1 0.5 0

The frequency and the duty cycle of the signal at Q will respectively be

(C)

(B) f (C) f

10

20 30 time (ms)

40 50

1

0.5

(D) f

0

0

10

20 30 40 time(ms)

10

20

50

(D) 1.5 current

EE-2009 6. An SCR is considered to be a semi-controlled device because (A) it can be turned OFF but not ON with a gate pulse (B) it conducts only during one half-cycle of an alternating current wave (C) it can be turned ON but not OFF with a gate pulse (D) it can be turned ON only during one half-cycle of an alternating voltage wave

1 0.5

0 0

8.

30 40 times(ms)

50

Match the switch arrangements on the top row to the steady-state V-I characteristics on the lower row. The steady state operating points are shown by large black dots.

The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50Hz voltage source. Under ideal conditions the current waveform through the inductor will look like

(a)

(b)

+

+

is

0.1 V

0

1.5 5

current

(A) f

is

(I)

10 sin100 T

(II)

(A) Vs

1.5 current

7.

Power Electronics

Vs

1

0.5 0

0

10

20 30 time(ms)

40

50

th

th

th


GATE QUESTION BANK

(c)

Power Electronics

(d) T

L

15V

+

(A) 10 s (B) 50 s

+ i

is

(III)

(IV) Vs

Vs

Codes: a (A) I (B) II (C) IV (D) IV

b II IV III III

c III I I II

d IV III II I

i

~ (i), (ii) and (iii)

(A) (B) (ii), (iii) and (iv) (C) (ii) and (iii) (D) (i) and (iv) 13.

EE-2013 11. Thyristor T in the figure below is initially off and is triggered with a single pulse of

)

b

b

EE-2012 10. The typical ratio of Latching current to holding current in a 20 A thyristor is (A) 5.0 (C) 1.0 (B) 2.0 (D) 0.5

and C =(

(C) 100 s (D) 200 s

EE-2014 12. Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches can block voltages of either polarity applied between terminals ‘a’ and ‘b’ when the active device is in the OFF state? a a a a

EE-2011 9. Circuit turn-off time of an SCR is defined as the time (A) taken by the SCR to turn off (B) required for the SCR current to become zero (C) for which the SCR is reverse biased by the commutation circuit (D) for which the SCR is reverse biased to reduced its current below the holding current

width 10 s. It is given that L =(

C

ii

b

b

~

iii

~

iv

~

A diode circuit feeds an ideal inductor as shown in the figure. Given V 100 sin t V, where 100 rad/s, and L = 31.83 mH. The initial value of inductor current is zero. Switch S is closed at t = 2.5ms. The peak value of inductor current i (in A) in the first cycle is________ t 2.5 ms i

)

S

. Assuming latching and

holding current of the thyristor are both zero and the initial charge on C is zero, T conducts for

V

th

~ th

th


GATE QUESTION BANK

Power Electronics


[Ans. A]

Anode current i i 0.02 5 1 e Let minimum pulse width is T To turn on ia ≥ latching current ⇒ 0.02 5 1 e ) 50m 0.05 T 150 sec

R2

R1 C VS

Th2

Th1

In this type of commutation, a thyristor carrying load current is commutated by transferring its load current to another incoming thyristor. Firing of SCR Th1 commutates Th2 and subsequently, firing of SCR Th2 would turn – off Th1. Circuit turn – off time t for Th t R ln 2 and circuit turn – off timet for Th t R ln 2 as R R 50 t t R cln 2 Safety margin = 2 So, R ln2 2t 2 50 10 c 2.88 50 ln2

[Ans. C] 10Ω

D1

V

20 0.5H

V2

V1

D2 + 200V

VCE

When the pulses are applied to the base of the transistor. Transistor operates in ON state. So, the forward voltage drop in transistor V 1 V. V 10 V 10 1 9V 1 V V ( ) V 9V [turn ratio 1 1] 1 is forward biased and voltage drop in diode V =1 V is reversed biased and acts as open circuit. Capacitor behaves as open circuit for dc voltage. Forward voltage drop of gate cathode junction V 1V Voltage drop across resistor R. V V V V 9 1 1 7V To ensure turn – ON of SCR, V 7 R 47 150m

iR R2=5K

R

K R

iA iL

L

G

ig

PT

+10V

[Ans. B]

V=100

2.

3.

Current through 5 k resistor V 100 i R 5 10 20m 0.02 Current through inductor V ⁄ i (1 e ) R 100 . (1 e ) 20 5 1 e th

th

th


GATE QUESTION BANK

4.

[Ans. A]

7.

[Ans. C] Frequency of the voltage source, f = 50 Hz

RL = 1 L

+ iL VT

Time period, T =

voltage, 0 < t< , energy is stored in the

Forward voltage drop of SCR during ON – state VT=1 V I C + V V di dt

Ri

V

inductor and current increases. During negative half cycle of the source voltage, I

0

8.

I

I

0.15

⇒

V V

V I

I

⇒

V

V

I In Parallel V

V

∴ Switch is ON if either T or impressed are ON and Switch is OFF, if both T or impressed D are OFF. I

T

1 T

T, current decreases and

[Ans. C]

[Ans. A] f

t

energy stored in the inductor is delivered to source.

di i 1 0 dt ⇒i 199(1 e ⁄ . ) Gate pulse width required = time taken by i to rise upto T ⇒ 250 10 199(1 e . ) T 188.56 s Width of the pulse = T = 188.56 s Magnitude of voltage = V = 10 V Voltage second rating of PT VT = 10 188.56 s 1885.6 v-s 2000 v-s 5.

20 ms.

During positive half cycle of the source E=200V

⇒ 200

Power Electronics

T

T

T

V 6.

[Ans. C] SCR can only be turn on by applying gate pulses but it cannot be turnoff by the same. Once the thyristor is on, and its anode current is above the latching current level the gate loses control. It can be turned off only by reducing the anode current below holding current. Only triggering can be done through gate but commutation can’t be done through gate gate loses control after the SCR turns on.

9.

[Ans. C] The turn – off time provided to the thyristor by a circuit is called circuit turn off time. It is defined as the time between the instant anodes current becomes zero and the instant reverse voltage due to the circuit reaches zero.

th

th

th


V

GATE QUESTION BANK

10.

11.

[Ans. B] For medium power thyristors of rating 6 A to 60A the ratio of the latching current to holding current is 1.5 to 2. [Ans. C] 1 f √ 1 f

T

1

i

V

at t i

1 100

Power Electronics

(cos ( ) cos t) 4 10ms is maximum V 1 100 10 ( cos ( )) 1000 √2 100 1 0.707 100 31.83 10 17.08

100 s

12.

[Ans. C] In (i) and (iii), when switch are reverse biased, diode in parallel will start conducting so not able to block the reverse voltage

13.

[Ans. *] Range 16.6 to 17.4 100 rad s 2 T 20ms 100 V

2ms

10ms

2ms

10ms

20ms

i

Start at which current is maximum at t = 10 ms di V sin t dt V sin t di ∫ dt i at t 0 k

V

cos t

k

2.5 ms i=0 V 100 2.5 cos ( ) 1000 V cos ( ) 4

k

th

th

th


GATE QUESTION BANK

Power Electronics

Phase Controlled Rectifier EE-2006 1. A 3-phase fully controlled bridge converter with freewheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 600. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be (A) IDF = 0.867; IPF = 0.828 (B) IDF= 0.867; IPF = 0.552 (C) IDF = 0.5; IPF = 0.478 (D) IDF = 0.5; IPF = 0.318 2.

A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 2 Ω as shown in figure. The SCRs are triggered by a constant dc signal. If SCR 2 gets open circuited, then what will be the average charging current?

i

V

0

SCR2

SCR3

SCR4

3.

(C) 11.9 A (D) 3.54 A

A single-phase half wave uncontrolled converter circuit shown in figure. A 2-winding transformer is used at the input for isolation. Assuming the load current to be constant and v V sin t the current waveform through diode will be

2

0

4.

2

A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. if the solar cell resistance is 0.5 , then each thyristor will be reverse biased for a period of (A) 125 (C) 60 (B) 120 (D) 55

50Hz

(A) 23.8 A (B) 15 A

2

0

Battery SCR1

2

0

200V

230V

V

EE-2007 5. A single-phase fully controlled thyristor bridge ac-dc converter is operating at a firing angle of 25 and an overlap angle 10 with constant dc output current of 20 A. The fundamental power factor (displacement factor)at input ac mains is (A) 0.78 (C) 0.866 (B) 0.827 (D) 0.9

th

th

th


GATE QUESTION BANK

6.

7.

A Single phase fullwave half-controlled bridge converter feeds an inductive load. The two SCRs in the converter are connected to a common DC bus. The converter has to have a free-wheeling diode. (A) because the converter inherently does not provide for free-wheeling (B) because the converter does not provide for free-wheeling for high values of triggering angles (C) or else the free-wheeling action of the converter will cause shorting of the AC supply (D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR

EE-2008 9. A single-phase half controlled converter shown in the figure is feeding power to highly inductive load. The converter is operating at a firing angle of 60°.

~

v

~

If the firing pulses are suddenly removed, the steady state voltage (v0) waveform of the converter will become (A) V 0

In the circuit of adjacent figure the diode connects the ac source to a pure inductance L. D

AC

Power Electronics

t

(B)

V 0

Pure L

3

4

3

2

t

3

4

3

2

t

2

t

(C)

The diode conducts for (A) 90 (C) 270 (B) 180 (D) 360

V 0

8.

2

A three-phase fully-controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power 420 V dc to a three phase, 415 V(line), 50Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is (A) 119.05 A (C) 68.73 A (B) 79.37 A (D) 39.68 A

(D) V 0

10.

th

A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30 . The approximate Total harmonic Distortion th

th


GATE QUESTION BANK

(%THD) and the rms value of fundamental component of the input current will respectively be (A) 31% and 6.8 A (C) 66% and 6.8 A (B) 31% and 7.8 A (D) 66% and 7.8 A

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10 Ω. 13.

11.

A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current. If the triggering angle is 30 , the input power factor will be (A) 0.65 (C) 0.85 (B) 0.78 (D) 0.866

Power Electronics

14.

EE-2010 12. The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0 , the dc output voltage of the converter is 300V. What will be the output voltage for a firing angle of 60 , assuming continuous conduction?

The maximum battery will be (A) 14 A (B) 40 A

current

through

the

(C) 80 A (D) 94 A

The kVA rating of the input transformer is (A) 53.2 kVA (C) 22.6 kVA (B) 46.0 kVA (D) 19.6 kVA Common Data for Questions 15 and 16 The input voltage given to a converter is v 100 √2 sin 100 t V The current drawn by the converter is i

(10√2 sin (100 t

) 3 5√2 sin 300 t

4

2√2 sin 500 t 6 ) 15.

The input power factor of the converter is (A) 0.31 (C) 0.5 (B) 0.44 (D) 0.71

16.

The active power drawn by the converter is (A) 181 W (C) 707 W (B) 500 W (D) 887 W

V

(A) 150 V (B) 210 V

(C) 300 V (D) 100 V

EE-2011 Statement for Linked Answer Questions 13 and 14 A solar energy installation utilizes a three-phase bridge converter to feed energy into power system through a transformer of 400 V/400 V, as shown below. Filter Choke

~

EE-2012 17. A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle  and the load current is continuous. The fraction of cycle that the freewheeling diode conducts is (A) 1 2 (C) ⁄2 ⁄ (B) 1 (D) ⁄

Battery

th

th

th


GATE QUESTION BANK

EE-2014 18. The figure shows the circuit of a rectifier fed from a 230-V (rms), 50-Hz sinusoidal voltage source. If we want to replace the current source with a resistor so that the rms value of the current supplied by the voltage source remains unchanged, the value of the resistance (in ohms) is_________ (Assume diodes to be ideal.)

~

10

230V 50 z

19.

~

20.

21.

A single-phae SCR based ac regulator is feeding power to a load consisting of 5 resistance and 16mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are (A) 30 and 46 A (C) 45 and 23 A (B) 30 and 23 A (D) 45 and 32 A

22.

The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 µs is applied to the SCR. The maximum value of R in to ensure successful firing of the SCR is___ S R

The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10Ω and an inductance 0.05 connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70 V is __________ 325 sin 314t V

Power Electronics

500 R

100V 200m

oad

A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply (v ) to the bridge is a sinusoidal source. Triggering angle of the bridge converter is 30 . The input power factor of the bridge is _________

23.

A three-phase fully controlled bridge converter is fed through star-delta transformer as shown in the figure. R

i V

~

1

Load

th

th

th


GATE QUESTION BANK

24.

The converter is operated at a firing angle of 30 . Assuming the load current ( ) to be virtually constant at 1 p.u. and transformer to be an ideal one, the input phase current waveform is

Power Electronics

2k 3 2

0

2 3k 2 3k 1 3k 0

2

2k 3 k 3 0

2

2

0


[Ans. C] Load current is constant 10 In 3-ϕ full converter with free-wheeling diode, input displacement factor (IDF) cos cos 60 0.5 R S value source current 2 3

√

1

√

2.

[Ans. C] Vi E v

2 3

i0

nth

RMS value of harmonic 4 n sin 3 √2n RMS value of fundamental current 2√2 √6 sin 60

average current 1 ∫ 2 R

Current distortion factor (CDF) √6 3

1 3 ( √ ) 2

∴

avg

V sin wt

1 [2V cos 2 R

]

2

1 2

0.955

2 [2 (230

√2) cos

Input power factor (IPF) where 0.955 0.478

d

0.5

sin sin

(

avg

1 2

2

2

]

)

V

200

230 √2 0.66 rad

38 ∴

(

200

[2√2 200

)

230 cos 38 2

0.66 ]

11.9 th

th

th


GATE QUESTION BANK

3.

[Ans. C]

⇒

V

440√2 cos

340

⇒ 125 Therefore, each thyristor will be reverse biased for a period of 55

t

5.

V

3

Power Electronics

[Ans. A] Firing angle= 25 Overlap angle = 10 DC output current V [cos cos

t

i

]

l

230 √2 2 50 [cos 25 cos 25 ∴ 0.0045 Average output voltage 2V cos v

t

∴ 20

i l t i i t

2 or 0 t Diode is forward biased and conduct. For t 2 Diode becomes reverse biased and Diode gets forward biased and starts conducting. As load current is constant current through Diode i can be drawn as shown in the figure. 4.

[Ans. D] Solar cell emf E = 350 V DC current 20 Solar cell resistance R 0.5 V Voltage across inverter R 350 20 0.5 340 V The bridge acts as inverter, Output voltage of 3 ϕ fully controlled bridge 3V V cos 3V

cos

230√2 cos 25 3.14 2 3.14 50

10 ]

4.5 3.14

10

20

187.73 9 178.74 V Displacement factor= 178.25 20 230 20 0.78 6.

[Ans. C] T2

T1

Vs

~

I0

V0

D1

D2

1-ϕ full wave half controlled bridge converter without free-wheeling diode is shown in the figure.

340

th

th

th


GATE QUESTION BANK Vs

9. 2

[Ans. A]

t

Vs

~

V0

t

D1

D2

2

Single phase half controlled converter As load is highly inductive, it means load current is continuous and almost constant.

At t T is fired and T1 starts conducting load current flows through T1D1 for t . At t V becomes negative and D1 gets reverse biased and D2 is forward biased. So during t free wheeling action takes place through T1 and D2 and output voltage become zeros. At t load T2 is triggered and load current is transferred from T1 to T2. So, during t 2 T conducts. At t It may be possible that load current is not transferred completely from T1 to T2, and T1 and T2 may be conducting simultaneously which results in short circuit of the supply for short direction. 7.

[Ans. D] Inductor current is positive for 360 , hence the diode conducts for whole of the period, i.e, 360 .

8.

[Ans. C] Let DC link current = DC voltage applied to the inverse V 420 Power fed to the inverter P V 50 k ⇒ 420 50 10 119.05 Current through each thyristor flow for period of 2 3 So, rms current of thyristor. √ √3

2

2 +

4 5

t

3

V0

T1D1 T1D2 T1D1 T1D1 T2D2 T2D1 T1D2 T2D1

t

no firing pulses

T1 is triggered

triggered

T1 is T2 is

During,  0 t freewheeling by T V 0  t T1 D1 conducts V V  t Freewheeling by T1D2, V0=0  t 2 T2D2 conduct , V0 = Vs T1 is again triggered at 2 So during 2 t 3 T , conducts. Now, if firing pulses are removed after T1 is triggered at 2 . At t 3 no firing pulse will be available to trigger T2. So, load current will flow through T1D2 as load current is continuous. So during 3 t 4 TD2 will continue to conduct and V0 =0. At t 4 D1 will become forward biased and D2 will become reverse biased. So, D1 will conduct and T1 is already conducting.

d t

∫

119.05

Vs

triggered

1

√3

T2

T1

V0

T2 D T1D1 T1D T2D2

Power Electronics

68.73

th

th

th


GATE QUESTION BANK V

So load current will flow through T1 D1 and V0 = Vs. 10.

[Ans. B] Load current = 10 RMS value of total source current √

2 3

10√

2 3

Supply current i fourier series i t

t V

8.165 2

can be expressed by

4 n sin sin n t n 3

∑

4

n

t i

t

10

√3 7.8 2 √2 Total Harmonic distortion √(

)

√(

8.165 ) 7.8

1

Average output voltage 1 V ∫ V sin t. d

100

cos

V

Average output current = l (constant) RMS value of supply current = l l RMS value of supply voltage

+ T

~

V T

V

√

Input power factor Power delivered to load nput V (√3 )V V V V ⁄√2.

T

+ V

√3

V

[Ans. B]

i

t

30

100

31 11.

2V

⇒V for

1

t

i

RMS value of fundamental current, 4 sin 3 √2 ⇒

Power Electronics

L O A D

T

√6

12.

0.78

[Ans. A] cos V

=

V

th

V = 300 V

th

. cos60

300.

th

150 V


GATE QUESTION BANK

13.

[Ans. B] Filter

~

p. f

Choke Battery(E)

14.

2 3

40√

2 3

17.

[Ans. D] V

0

18.

32.66

10

over

∴ raction [Ans. *] Range 23 to 23 RMS value of current supplied by source 10 10

√3V 32.66

2

Freewheeling diode conducts for 2 complete cycle

KVA rating of the input transformer √3 400 22.62 V

0.44

[Ans. B] Active power will be drawn by converter only due to fundamental component. Therefore, Active power = V cos ϕ 100 10 cos 60 Active power = 500 watts

[Ans. C] RMS value of supply current in case of 3 – ϕ bridge converter √

10 cos 60 11.35

16.

Average output voltage of the converter 3V V cos The converter acts as line commutated inverter and for such mode 90 and V is negative. Therefore battery supplies energy to AC system. So, current through battery 400 V R for V 0 or 90 Maximum current flow through battery 400 40 10

Power Electronics

V

T 2

T t

15.

[Ans. B] Voltage applied (v) = 100√2 sin 100 t Current resulted 10√2 sin (10 t

) 5√2 (300 t 3 2√2 sin (500 t

10

4 )

RMS value of current supplied when resistor is there

)

V

The current flown into converter consists of fundamental, 3rd harmonic and 5th harmonic components. In the case of converter drawing non sinusoidal component of current then p.f will be written as Input power factor =

t

V

10 √2R V 230√2 230√2 ⇒ 10 √2R R 23

cos ϕ

10 √10

5

2

R

11.35 th

th

th


GATE QUESTION BANK

19.

[Ans. *] Range 69 to 70 Freewheeling diode is there so after freewheeling diode will conduct and energy stored in inductor will dissipate through FD and R. V

2√2

cos 30

2√2

21.

Power Electronics

√3 2

√6

[Ans. C] v sin z

t

0.78

ϕ

t

i

2

~ 1 V ∫ V sin t d t 2 V 1 cos 2 1 cos 325 70 2 3.14 cos 0.353 69.33 20.

[sin

i

t

ϕ

sin f

ϕ e v sin z

ϕ⇒i

+

t

ϕ

v v Voltage across device is zero ϕ

[Ans. *] Range 0.74 to 0.82 V 2

tan

(

tan

(

) 2

50 16 m ) 5

45.15 v v z z√2

i

230 √ 5 23

2

22.

output power V

output power V 1 V ∫ V sin t d t V 2V V

2V

rom figure V

cos t |

i t

cos

40 ∴R

cos

nput power factor

50

16m

[Ans. *] Range 6055 to 6065 Given data 40 m t 50 s find R

Highly inductive load So is continuous nput power factor

2

V *1 R

100V

e +

100 [1 500 6060.83

R

V R

10

e

]

200m 500 2m 5

100 R

cos √ th

th

th


GATE QUESTION BANK

23.

Power Electronics

[Ans. B] V

V

V

V

V

1 3

i

i

k 3

i

input waveform i 2k⁄3 i i ⁄

k 3

th

th

th


GATE QUESTION BANK

Power Electronics

Choppers EE-2006 Statement for Linked Answer Questions 1 and 2 A voltage commutated chopper operating at 1 kHz is used to control the speed of dc motor as shown in figure. The load current is assumed to be constant at 10 A. +

1 F

2.

M

IL = 4A L

2mH

The average output voltage of the chopper will be (A) 70 V (C) 35 V (B) 47.5 V (D) 0 V

EE-2007 3. The circuit in the figure is a current commutated dc – dc chopper where, Th is the main SCR and Th is the auxiliary SCR. The load current is constant at 10 A. Th is ON. Th is triggered at t=0. Th is turned OFF between.

(A) 10 V, 2A (B) 10 V, 8 A

Th Load 10

25.28µH

Load V0

(C) 40 V, 2 A (D) 40 V, 8 A

EE-2009 5. In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on T is 50 V s what should be the theoretical minimum value of C1? Assume current ripple through L0 to be negligible. T T 8Ω

(A) 0.2 (B) 0.02

Th

(A) (B) (C) (D)

D

‘S’

20V

The minimum time in sec for which the SCR M should be on is (A) 280 s (C) 70 s (B) 140 s (D) 0 s

230V

ID

A

V = 250V

1.

EE-2008 4. In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous. The average voltage across the load and the average current through the diode will respectively be

(C) 2 (D) 20

EE-2010 6. The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a _______ A

0 µs t 25 µs 25 µs t 50 µs 50µs t 75 µs 75 µs t 100 µs

V

th

L

P

B

th

V

th


GATE QUESTION BANK 100

(A) step-down chopper (buck converter) (B) half-wave rectifier (C) step-up chopper (boost converter) (D) full-wave rectifier EE-2011 7. A voltage commutated chopper circuit, operated at 500 Hz, is shown below. M 0.1µF

A

i

i

Power Electronics

12V Q

20 470

9.

The average source current in Amps in steady – state is (A) 3/2 (C) 5/2 (B) 5/3 (D) 15/4

10.

The PEAK-TO-PEAK source current ripple in amps is (A) 0.96 (C) 0.192 (B) 0.144 (D) 0.288

10

i

200V

LOAD

1mH

If the maximum value of load current is 10 A, then the maximum current through the main (M) and auxiliary (A) thyristors will be (A) i 12 and i 10 (B) i 12 and i 2 (C) i 10 and i 12 (D) i 10 and i 8 EE-2012 8. In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50 . Given that ∆ic is 1.6 A peak-to-peak and I0 is 5 A dc, the peak current in S is S

EE-2014 11. Figure (i) show the circuit diagram of a chopper. The switch S in the circuit in figure (i) is switched such that the voltage V across the diode has the wave shape as shown in figure (ii). The capacitance C is large so that the voltage across it is constant. If switch S and the diode are ideal, the peak to peak ripple (in A) in the inductor current is___________ S 100V

1m V

oad igure i

V

i 100V

V

24 V

R

(A) 6.6 A (B) 5.0 A

EE-2013 Common Data Questions 9 and 10 In the figure shown below, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 kHz, with a duty ratio of 0.4. All elements of the circuit are assumed to be ideal.

0.05

0

(C) 5.8 A (D) 4.2 A

0.1

0.15

0.2

t ms

igure ii

12.

th

A step-up chopper is used to feed a load at 400 V dc from a 250 V dc source. The inductor current is continuous. f the ‘off’ time of the switch is 20 µs, the switching frequency of the chopper in kHz is_____________

th

th


GATE QUESTION BANK

Power Electronics


2.

V

D

5.

[Ans. A] 0.8

c +

LOAD

Th

+ 230 V

L

vc

V

V √ sin sin

t - i

0

V V and i . As i tends to reverse. Th is turned off. When V V , right hand plate has positive polarity , resonant current i now builds up through C, L, D and Th . As this current of Th , net forward current i i begins to decrease. Finally when i in the reversed direction attains the value I0 , im is reduced to zero and Thm is turned off. i i sin ∆t 0 1 ∆t sin ( )

dv dt

50V

sec

V 0.8 100 ⇒ 80 V V 80 i 10 R 8 During commutation of main SCR T load current will be flown through capacitor. dv ∴ i dt 10 10 ⇒ ⇒ 0.2 f 50

t

After half a cycle of i ,t

100 V and

V

At t =0 v V i 0 and i . At t =0, Th is triggered, a resonant current i designs to flows from C throughTh , L and back to C. This resonant current is given by i

√

[Ans. C] When S is ON, V = 0V and = 0A When S is off, V 20 V and = 4A v 20 ∴ V 40V 1 d 1 0.5 0.5 × 4 = 2A

I0

im

∆t

4.

[Ans. C] Th

t

√10 25.28 sec 50 sec Option (C) is correct. Since commutation of Thm starts from t1=50 sec

250

= 35V

t

t

[Ans. C] V

3.

t

[Ans. B] Minimum time for which SCRm should be ON is same as time for which C charges from + 250 V to 250V. T= √ 140 sec. √20 10

6.

[Ans. A] B and D are not possible options as the input is dc. P→ V

V (1

P →  V = V .e If L is large enough, V V when P → 0 when P → ∴ step down chopper

e ) where T= L/R

So, Th is turned off between th

th

th


GATE QUESTION BANK

7.

[Ans. A] When main thyristor (M) is turned on, an oscillatory current in the curcuit C, M, L and diode is set up and it is given by i t

V √ sin

10.

20

Given f = 250 k 1 So T 250 10

i Vin

T

∆

5

.

T

t

i. e. Peak – peak inductor current . .

T 0.192

11.

[Ans. C] Peak Current = I0 +

9.

470

We have to find out peak to peak source ripple current here ource current and inductor current are same peak to peak source ripple = peak to peak inductor current V T → uty ratio

0.1 10 1 10

2 Current through main thyristor i t sin t So, maximum value of i 10 2 12 When auxiliary thyristor (A) is turned on capacitor voltage applies a reverse voltage across main thyristor and main thyristor is turned off. The load current is now carried by c and auxiliary thyristor. Current through auxiliary thyristor i Maximum value of iA = maximum value of I0 = 10A

8.

12 V

t

V√

√

[Ans. C] Given circuit :

V

Peak value of current through capacitor

200

Power Electronics

5.8

[Ans. B] v V 1 12 20 V 1 0.4 20 1 20 Given that all are ideal P P ⇒ V 20 1 ∴ 5 3

[Ans. *] Range 2.49 to 2.51 For 0 t 0.05 ms V 100 V or 0.05 t 0.1 ms V V Using volt-sec, across inductor 100 V 0.05 V 0.05 V 50 Slope

50 Slope

0.05

V

th

0

50

0.1

di dt th

th


GATE QUESTION BANK

di dt

V

v ∫ dt

⇒i

Peak to peak current i 50 50

12.

0.05 0.05 10

V

Power Electronics

t t

10 10

2.5

[Ans. *] Range 31.0 to 31.5 V 1 V 1 400 1 250 1 8 8 5 3 8 5 T 20 S 8 T 32 s 10 f 31.25 k 32

th

th

th


GATE QUESTION BANK

Power Electronics

Inverters EE-2006 1. A single- phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each half cycle as shown in figure. It is found that the output voltage is free from 5th harmonic for pulse width 1440. What will be percentage of 3rd harmonic present in the output voltage (Vo3/Vo1max)? V

2d → 3

2 2

2 2d →

V

(A) 0.0% (B) 19.6%

(C) 31.7% (D) 53.9%

EE-2007 2. “Six OS Ts connected in a bridge configuration (having no other power device) MUST be operated as a Voltage Source nverter VS ”. This statement is (A) True, because being majority carrier devices, MOSFETs are voltage driven (B) True, because MOSFETs have inherently anti-parallel diodes (C) False, because it can be operated both as Current Source Inverter (CSI) or a VSI (D) False, because MOSFETs can be operated as excellent constant current sources in the saturation region. 3.

A Single –phase voltage source inverter is controlled in a single pulse-width modulated mode with a pulse width of 1500 in each half cycle. Total harmonic distortion is defined as √

THD =

× 100,

Where V1 is the rms value of the fundamental component of the output voltage. The THD of output ac voltage wave form is (A) 65.65% (C) 31.83% (B) 48.42% (D) 30.49% EE-2008 4. A 3-phase Voltage Source Inverter is operated in 180° conduction mode. Which one of the following statements is true? (A) Both pole-voltage and line-voltage will have 3rd harmonic components. (B) pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic (C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic (D) Both pole-voltage and line-voltage will be free from 3rd harmonic components. 5.

A single phase voltage source inverter is feeding a purely inductive load as shown in the figure. The inverter is operated at 50 Hz in 180 square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be________

0.1 H 200 V

I0

(A) 6.37 A (B) 10 A

(C) 20 A (D) 40A

EE-2009 6. The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs T T and T T . If the load is purely resistive, the theoretical th

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GATE QUESTION BANK

maximum output frequency obtainable will be T

T

0.1 + 10

10 A +

T

0.1

T

(A) 125 kHz (B) 250 kHz

Power Electronics

EE-2013 Statement for Linked Answer Questions 8 and 9 The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square-wave ac output voltage V across an R-L load. Reference polarity of v and reference direction of the output current i are indicated in the figure. It given that R = 3 ohms, L = 9.55mH.

(C) 500 kHz (D) 50 kHz

Q

Q

L

V

EE-2011 7. A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S to S are identical switches. 50H S z

S

i0

v

R

Q

Q

8.

In the interval when v0 < 0 and i0>0 the pair of devices which conducts the load current is (A) Q Q (C) (B) Q Q (D)

9.

Appropriate transition i.e., Zero Voltage Switching (ZVS)/Zero Current Switching (ZCS) of the IGBTs during turn-on/ turnoff is (A) ZVS during turn – off (B) ZVS during turn – on (C) ZCS during turn – off (D) ZCS during turn – on

S . .

S

S

S

The proper configuration for realizing switches S to S is (B) (A) A A

B

B

(C)

A

(D)

A

EE-2014 10. The figure shows one period of the output voltage of an inverter. should be chosen such that 60 90 . If rms value of the fundamental component is 50 V, then in degree is _____________ 100V

0

B

100V

100V

180

180 180

100V

B th

th

100V

th

360 360

t degree

100V


GATE QUESTION BANK

11.

Power Electronics

A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure. S i

O

V

S

oad

S

S

S S

2

S S 2

If the load current is sinusoidal and is zero at 0 2 . . , the node voltage V has the waveform

V

2

V

2

V

2

V

2

V

2

V

2 V

V

2

V V

2

2 2

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Power Electronics


[Ans. B] The output of inverter = v ∴ v ∑

v

4V sin nd sin n t sin n n

Here pulse width = 2d =144 ⇒ d 72 4V v sin 72 sin t for n 4V sin 3 3

v

sin 3

V V max

72

RMS value of fundamental component of output voltage. 1 4V V [ sin 75 ] 0.87 V √2 Total Harmonic Distortion

2

1

72 sin 3 t for n

√

T

3

20.6

sin 72

√

Option (B) is correct as it is nearer to 20.6%.

V

0.913 V 0.87V 0.87V

∑

V

V0

Vs

3 2 t 2d

(

2

d) 2

2

5.

√

1

4V n cos sin n 6

3 cos ∑

cos

t

n .. i 6 0

2V sin n t n

ii

[Ans. B] V0

vs

Vs

Pulse width = 2d = 150 ⇒ d 75

V

100

It is clear from eq. (i) as (ii) both check voltage and line voltage will be from 3rd harmonic components.

d 2d

100

V

[Ans. C]

2

V

[Ans. D] Line voltage and pole voltage of 3 – ϕ VSI operated in 180 conduction mode can be expressed by the fourier series as follows.

[Ans. B] As MOSFET have internally antiparallel diodes, MOSFET cannot withstand reverse voltage, due to which MOSFET cannot be as current source inverter (CSI).

or n

3.

V

31.83

4. 2.

4V n sin sin nd sin n t n 2

∑

2

-vs

t

i0

∫ V d

t

iP 2

t

iP

V√

2d

V√

5 6

0.913 V

Output voltage in fourier series.

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GATE QUESTION BANK

Load current increases linearly from IP to IP during 0 < t . Where Ip = Peak value of i0 1 ⇒ t sec 2 50 100 t t 0i and t t i di V dt During 0< t V V 200 V ⇒ 200 ⇒

6.

7.

0.1 .

t

0

Power Electronics

cannot be used in a CSI. So, a diode is added in series with the devices for reverse blocking.

8.

[Ans. D] V

i

/

200 t 0.1 2000 1 10 2 100

2

[Ans. B] The circuit shown in the figure is a single phase bridge auto sequential commutated inverter (1-phase ASCI) Thyristor pairs T T and T T are alternatively switches to obtain a nearly square wave load current. Two commutating capacitors, one in the upper half and the other in the lower half are connected as shown Diodes to are connected in series with each SCR to prevent the commutation capacitors from discharging into the load. The inverter output frequency is controlled by adjusting the period T through the triggering circuits of thyristors. The theoretical maximum output frequency obtainable 1 f 4R 1 4 10 0.1 10 250k z

Voltage and current waveform of a full bridge inverter Shaded portion shows the condition where V 0 i 0 Conducts the load current

9.

[Ans. D] Turn on points for thyristors Q Q Q Q are given ∴ A, B, C represents points where current cross over zero. Zero current switch (ZCS) during turn on.

10.

[Ans. *] Range 76.5 to 78.0 Fourier series a 2

f t

∑ a cos n t ∑ b sin n t

Fundamental component n = 1 ⇒ a cos t b sin t 1 a ∫ f t cos t d t

[Ans. A] Device used in current source inverter (CSI) must have reverse voltage blocking capacity. Therefore, device such as GTOs, power transistors and power MOSFETs th

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Power Electronics

100 0∫ cos t d t

∫

cos t d t

∫

⇒a b

1

∫

cos t d t

∫

cos t d t

∫

cos t d t 1

f t sin t d t

400

[8 cos

[1

R S value ⇒ cos 76.99

11.

∫

0

100 b

cos t d t

4]

2 cos ] 400

√2 0.225

[1

2 cos ]

50

[Ans. D] During to s s ON V So V 2 During to anti parallel diode of s and s will conduct to maintain the flow of load current in positive direction V V 2 uring to Antiparallel diode of s and s conducts to maintain flow of load current in negative direction V V 2 During to 2 s s conducts V so V 2

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Power Electronics

AC Voltage Regulators and Cycloconverters EE-2008 1. In the single-phase voltage controller circuit shown in the figure, for what range of triggering angle ( ), the output voltage (Vo) is not controllable? T1

S

50

+

~

Vs

EE-2012 Common Data for Questions 2 and 3 In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180o conduction mode. All the switching devices are ideal. S

V S

j50

S

R

S R

3-phase inverter

(A) (B) (C) (D)

R 20

V

V0

T2

S

00 < < 450 450 < < 1350 900 < < 1800 1350 < <1800

3-phase balanced load

2.

The rms value of load phase voltage is (A) 106.1 V (C) 212.2 V (B) 141.4 V (D) 282.8 V

3.

If the dc bus voltage Vd = 300V, the power consumed by 3-phase load is (A) 1.5 kW (C) 2.5 kW (B) 2.0 kW (D) 3.0 kW


[Ans. A] For 0 <

< ϕ, V is not controllable

where ϕ= tan (

2.

ϕ

tan

ϕ ϕ

tan 45

⁄R) 50 ( ) 50 1

[Ans. B] RMS Value of line voltage = VL = √ VS RMS value of phase voltage = =

3.

√

√

=

√

VS

x 300 = 141.42V

[Ans. D] P=3.

=3x

.

= 3000W

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GATE QUESTION BANK

Power Electronics

Applications of Power Electronics EE-2006 1. The speed of a 3 – phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be (A)

f

V

50 (D)

f

V

50 (C)

3.

A single-phase,230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12V dc) with the series current limiting resistor being 19.04 . The charging current is ______ (A) 2.43 A (C) 1.22 A (B) 1.65 A (D) 1.0 A

V

50 (B)

power factor of the ac mains at half the rated speed, is (A) 0.354 (C) 0.90 (B) 0.372 (D) 0.955

f

V

EE-2008 4. A 220 V, 20 A, 1000 rpm, separately excited dc motor has an armature resistance of 2.5 Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input dc voltage to the chopper is 250 V The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be _____________ (A) 0.518 (C) 0.852 (B) 0.608 (D) 0.902 5.

A 220 V, 1400 rpm, 40A separately excited dc motor has an armature resistance of 0.4 Ω. The motor is fed from a single phase circulating current dual converter with an input ac line voltage of 220 V (rms). The approximate firing angles of the dual converter for motoring operation at 50% of rated torque and 1000 rpm will be _______________ (A) 430, 1370 (C) 390, 1410 0 (B) 43 , 47 (D) 390,510

6.

A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent

f 50

EE-2007 2. A Three-phase, 440V, 50Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15kW, 1500rpm separately excited dc motor with a ripple free continuous current in the dc link under all operating conditions. Neglecting the losses, the

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GATE QUESTION BANK

circuit as shown in the figure. Assume that the load inductance is sufficient to ensure continuous and ripple free load current the firing angle of the bridge for a load current of I0 = 10A will be _____________ 2𝛀

230v, 50hz 150V

(A) 44o (B) 51o

Power Electronics

EE-2013 7. The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If La =0.1mH, Ra = 1 , neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is

(C) 129o (D) 136o

La, Ra 200 V

(A) 0.4 (B) 0.5

(C) 0.6 (D) 0.7


2.

[Ans. A] In induction motor ϕ V/f If ϕ is constant then V f (Linear relation) [Ans. A] V 440 V f specification. V 440 V P

50 z → 3

= 3.

.

0.354

[Ans. D] 19.04

230 V 50 Hz

phase Supply

15kw N 1500 r. p. m → Rating of motor If losses are neglected, then R=0 and V = E = 440 V If speed in the separately excited motor become half, then V = E =220 V Power P = 15 kw P= I0 E 15 10 ⇒ 34.09 440

V

12 V

N ) N 1 230 ( ) 4 57.5 V

sin

1 2 R

√2

57.5 V

V (

sin

Rms Source current = I0√ 34.09

. √

√3 27.83 Supply power factor at half rated speed

( ) V 12 ( ) 12.04 57.5 180 12.04 ∫

1 *V 2 R th

th

V sin t cos t

th

169.76 d t +


GATE Question Bank

1 2

R

[V [cos

7.

]

cos

[Ans. D] 20 V 150 130 V

] 1 2 19.04 *57.5√2 cos 169.76

cos 12.04

12 (169.76 180

2

12.04

V

220

rad/sec 2.5

10

130 =140 V 140 V 200

[Ans. B] At rated conditions,

Ra 20 1

N Ra 10 1

V

)+

1 4.

Power Electronics

N

0.7

20

 At required conditions, V 2.5 20

5.



× 250 =



=

50

= 0.608

[Ans. C] At rated conditions, Vt = Eb + Iara = Km m + Iara  Km 204 x 60 2800 Nm At required conditions, armature current

20

and N = 1000

rpm V

cos √



r

cos 20

∴ 6.

39 and

0.4

141

[Ans. C] 10 ; V 230√2 V E = 150 V; r 2Ω cos

r



√

 cos

=



cos

150 √

20 √

= 129°

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Electrical Machines

Transformer EE-2006 1. In transformers, which of the following statements is valid? (A) In an open circuit test, copper losses are obtained while in short circuit test, core losses are obtained (B) In an open circuit test, current is drawn at high power factor (C) In a short circuit test, current is drawn at zero power factor (D) In an open circuit test, current is drawn at low power factor 2.

3.

4.

Two transformers are to be operated in parallel such that they share load in proportion to their kVA ratings. The rating of the first transformer is 500 kVA and its pu leakage impedance is 0.05 pu. If the rating of second transformer is 250 kVA, then its pu leakage impedance is (A) 0.20 (C) 0.05 (B) 0.10 (D) 0.025 Common Data Question for 3 and 4 A 300 kVA transformer has 95% efficiency at full load 0.8 pf lagging and 96% efficiency at half load, unity pf. The iron loss (Pi) and copper loss (Pc) in kW, under full load operation are (A) Pc = 4.12, Pi = 8.51 (B) Pc = 6.59, Pi = 9.21 (C) Pc = 8.51, Pi = 4.12 (D) Pc = 12.72, Pi = 3.07 What is the maximum efficiency (in %) at unity pf load? (A) 95.1 (C) 96.4 (B) 96.2 (D) 98.1

EE-2007 5. In a transformer, zero voltage regulation at full load is (A) not possible (B) possible at unity power factor load

(C) possible at leading power factor load (D) possible at lagging power factor load 6.

A Single-phase 50 kVA, 250V/500V two winding transformer has an efficiency of 95% at full load, unity power factor. If it is reconfigured as a 500V / 750V autotransformer, its efficiency at its new rated load at unity power factor will be (A) 95.752% (C) 98.276% (B) 97.851% (D) 99.241% Common Data for Questions 7, 8 and 9 A three phase squirrel cage induction motor has a starting current of seven times the full load current and full load slip of 5% If an autotransformer is used for reduced voltage starting to provide 1.5 per unit starting torque, the autotransformer ratio (%) should be (A) 57.77% (C) 78.25% (B) 72.56% (D) 81.33%

7.

8.

If a star-delta starter is used to start this induction motor, the per unit starting torque will be (A) 0.607 (C) 1.225 (B) 0.816 (D) 1.616

9.

If a starting torque required then the current should be (A) 4.65 (B) 3.75

of 0.5 per unit is per unit starting (C) 3.16 (D) 2.13

EE-2008 10. Three single-phase transformers are connected to form a 3-phase transformer bank. The transformers are connected in the following manner

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GATE QUESTION BANK

Electrical Machines

(A)

a2 a1

a1 a1

a1

b2 a1

b1 a1

a1 a1 Primary

c2 c1 a1Secondary a1

a1

a1

ers 24V

2 2.5

t(s)

48V (B) e

The transformer connection will be represented by (A) Y d0 (C) Y d6 (B) Y d1 (D) Y d11

48 V 1

t(s)

2 2.5

24 V

(C) ers

11.

12.

It is desired to measure parameters of ⁄ , 2KVA, single-phase transformer. The Following wattmeters are available in a laboratory W1: 250 V, 10 A, Low Power Factor W2: 250 V, 5 A, Low Power Factor W3: 150 V, 10 A, High Power Factor W4: 150 V, 5 A, High Power Factor The wattmeters used in open circuit test and short circuit test of the transformer will respectively be (A) W1 and W2 (C) nd (D) nd (B) nd The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure.

+

p

e

r 100

+

48V 24V

0

1

2

2.5

t(s)

(D) ers 0

1

2 2.5

t(s)

24V 48V

EE-2009 13. The single phase, 50Hz, iron core transformer in the circuit has both the vertical arms of cross sectional area 20cm2 and both the horizontal arms of cross sectional area 10cm2. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will

e

200

q

s

ф( b) 0.12

1 0 1 The induced 2 2. winding as a 5 the form t( s)

2

2.5

(A) (B) (C) (D)

t(s)

double remain same be halved become one quarter

emf (ers) in the secondary function of time will be of

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GATE QUESTION BANK

Common Data for Question 14 and 15 a A

Electrical Machines

i(t) 30 ms

B

t 5ms

10ms 15ms 20ms

25 ms 30ms 10A

The star-delta transformer shown above is excited on the star side with a balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition.

15.

0

c

C N

14.

10A

b

With both S1 and S2 open, the core flux waveform will be (A) A sinusoid at fundamental frequency (B) Flat-topped with third harmonic (C) Peaky with third-harmonic (D) None of these

16.

The peak voltage across A and B, with S open is (A) (B) 800V

17.

25 ms

(C) (D)

If the waveform of i(t) is changed to i(t)=10 sin (100 πt) , the peak voltage across A and B with S closed is (A) 400V (C) 320V (B) 240V (D) 160V

With S2 closed and S1 open, the current waveform in the delta winding will be (A) a sinusoid at fundamental frequency (B) flat-topped with third harmonic (C) only third-harmonic (D) none of these

Statement for Linked Answer Questions: 18 & 19 A C

Common Data Questions: 16 and 17 The circuit diagram shows a two winding, lossless transformer with no leakage flux, excited from a current source, i(t), whose waveform is also shown. The transformer has a m gnetizing induct nce of 4 /π mH.

The figure above shows coils 1 and 2, with dot markings as shown, having 4000 and 6000 turns respectively. Both coils have a rated current of 25A. Coil 1 is excited with single phase, 400V, 50Hz supply

1:1

Coil 1 B

18.

A S

i(t)

Coil 2

Ω

B

th

D

The coils are to be connected to obtain a single phase, 400/1000V, auto transformer to drive a load of 10kVA. Which of the options given should be exercised to realize the required auto transformer? (A) Connect A and D; Common B (B) Connect B and D; Common C (C) Connect A and C; Common B (D) Connect A and C; Common D

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GATE QUESTION BANK

19.

In the autotransformer obtained in Question 18, the current in each coil is (A) Coil 1 is 25 A and Coil 2 is 10 A (B) Coil 1 is 10 A and Coil 2 is 25 A (C) Coil 1 is 10 A and Coil 2 is 15 A (D) Coil 1 is 15 A and Coil 2 is 10 A

Electrical Machines

(A) (B) (C) (D)

( +j )Ω (0.866 – j . ) Ω ( .866 + j . ) Ω ( +j )Ω

EE-2011 22. A single-phase air core transformer, fed from a rated sinusoidal supply, is operating at no load. The steady state magnetizing current drawn by the transformer from the supply will have the waveform (A) i

EE-2010 20. A single-phase transformer has a turns ratio of 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1A, and the secondary current is 1A. If core losses and le k ge re ct nce’s re neglected, the primary current is

t

1A 1:2

~

i

(B)

(A) 1.41A (B) 2A 21.

(C) 2.24A (D) 3A

t

A balanced star-connected and purely resistive load is connected at the secondary of a star-delta transformer as shown in figure. The line-to-line voltage rating of the transformer is 110V/220V. Neglecting the non-idealities of the transformer, the impedance ‘Z’ of the equivalent star-connected load, referred to the primary side o the transformer, is: 110/220V R

(D)

r

b

i

4Ω

4Ω

t

v

B R Z Z

B

t

4Ω

Y

Y

(C) i

Z

EE-2012 23. A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W, at rated voltage and frequency, on no load. A second transformer has a th

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GATE QUESTION BANK

core with all its linear dimensions √ times the corresponding dimensions of the first transformer. The core material and lamination thickness are the same in both transformers. The primary windings of both the transformers have the same number of turns. If a rated voltage of 2 kV at 50 Hz is applied to the primary of the second transformer, then the no load current and power, respectively, are (A) 0.7 A, 77.8 W (C) 1A, 110 W (B) 0.7 A, 155.6 W (D) 1A, 220 W EE-2014 24. The core loss of a single phase, 230/115V, 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage, variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively, (A) 508 W and 542 W. (B) 468 W and 582 W. (C) 498 W and 552 W. (D) 488 W and 562 W. 25.

26.

27.

For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively, are (A) 10 and 21 (C) 21 and 10 (B) 10 and 21 (D) 21 and 10

28.

An open circuit test is performed on 50 Hz transformer, using variable frequency source and keeping V/f ratio constant, to separate its eddy current and hysteresis losses. The variation of core loss/frequency as function of frequency is shown in the figure

f

f(Hz)

sin ( t) y

(C) (D)

( /Hz)

The hysteresis and eddy current losses of the transformer at 25 Hz respectively are (A) 250 W and 2.5 W (B) 250 W and 62.5W (C) 312.5 W and 62.5 W (D) 312.5 W and 250 W

x

sin( t) , 4Ω sin( t), Ω

A single phase, 50 kVA, 1000V/100 V two winding transformer is connected as an autotransformer as shown in the figure.

The kVA rating of the autotransformer is _____________

Assuming an ideal transformer, the thevenin’s equiv lent volt ge nd impedance as seen from the terminals x and y for the circuit in figure are Ω

(A) (B)

Electrical Machines

sin( t) , Ω sin( t) , . Ω

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29.

The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.

(A) 5mH, 20mH and 40mH (B) 5mH, 80mH and 40mH (C) 25mH, 10mH and 20mH (D) 45mH, 30mH and 20mH

Hz c source

Ω

~

Z

31.

Assuming the transformer to be ideal, the value of the reactance X to improve the input power factor to unity is _____ 30.

Electrical Machines

The parameters measured for a 220V/110V, 50 Hz, single-phase transformer are: Self-inductance of primary winding = 45mH Self-inductance of secondary winding = 30 mH Mutual inductance between primary and secondary windings = 20 mH Using the above parameters, the leakage ( , ) and magnetizing ( ) inductances as referred to primary side in the equivalent circuit respectively, are

For a specified input voltage and frequency, if the equivalent radius of the core of a transformer is reduced by half, the factor by which the number of turns in the primary should change to maintain the same no load current is (A) 1/4 (C) 2 (B) 1/2 (D) 4


[Ans. D]

Circuit model in Open-Circuit test In open-circuit test, the transformer draws only exciting current. The exciting Current is only magnetizing in nature and is proportional to the sinusoidal flux and in phase with it, this is represented by Lagging the induce emf by . However the presence of eddy-currents, and hysteresis, both demand the flow of active power into the system and as a consequence the exciting current has another component in phase with . Thus the exciting current lags the induced emf by an angle slightly less than making power factor very low

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2.

Electrical Machines

100 A

[Ans. B] For generator

200 A

250 V

500 V

p.u le k ge imped nce ∝ 500 kVA × 0.05 = 250 kVA × x ∴x 3.

.

200 A

.

[Ans. C] Efficiency η

750 V (

%

96% =

300 A ) .

(

) . . (

. (

… (i)

) )

.

500 V

… (ii) , KVA rating of transformer

From, equation (i) p + p = 12.63 From equation (ii) 0.25 p + p = 6.25 From above two equations p 8. , p 4.

For auto transformer, η 7.

4.

= 98.276%

[Ans. C]

[Ans. B] x=√

.

%

.6 6

. .

η 5.

.

.

.

. /

= 96.2%

[Ans. C] Voltage regulation = I( cos + sin ) (For lagging power factor) Voltage regulation = I ( cos sin ) (For leading power factor) Voltage regulation = I( cos sin ) =0 t n

. ( ) x = 78.25% 8.

x

.

[Ans. B] . / ( )

9.

.

= 0.816

[Ans. C] . /

So zero voltage regulation is possible for leading power factor When t n

x

f

( ) . . ∴ p.u st rting current 10.

. 6

[Ans. B] A2

6.

[Ans. C] η ∴w

.

A

= 0.95 = + w = 2.631 C

B B2

C2

th

th

th


GATE QUESTION BANK a2 c c2

Electrical Machines

d dt d* . 4( . dt

e

a

c

t)+

48

a

N e

b b

48

b2 ∴ The possible connections is Yd

2

t(s)

2.5

4 11.

[Ans. D] For open circuit test [on LV side ] The current is of 4 to 8% of full load current ≅ .4 nd is of ow power f ctor ≅ . ∴ W2 is required

13.

oil

For short circuit test [on HV side] The current is full lo d current ≅ 8. The power f ctor is high ≅ . to .6 ∴ 3 is required 12.

[Ans. A]

oil

Reluctance of the magnetic circuit remains same in both the cases other windings are wound on horizontal arms or vertical arms. Self-inductance of a coil

[Ans. B] Induced emf (e ) in the secondary winding is given by d e where dt During O t The flux ( ) increases linearly with time . t d e dt d( . t) dt 4 During t is const nt . d e dt d( . ) dt uring t . decre ses line rly with time . 4( . t)

elut nce ∝ When the coils are wound on vertical arms. ( ) ∝( ) nd ( ) ∝ ( ) ross-section l re cm Assuming square cross-sectional area. Side of square √ Let length of the coil = L No. of turns 4 ( ) ∝(

∝

√ )

∝

Similarly, ( ) ∝ utu l induc t nce th

th

th


GATE QUESTION BANK

When S is closed, the same induced voltage appears across the Resistive load ∴ pe k volt ge cross & 4

m ∝ √( ) ( ) ∝ √(

)(

Electrical Machines

)

m ∝ / Similarly, when the coils are placed on horizontal arms

18.

[Ans. A] Hint: Polarity should be such that the voltages of two coils are additive. 10

m m m m m There, mutual inductance gets double when windings are placed on horizontal arms. 14.

15.

16.

[Ans. B] For sinusoidal excitation, the flux is a flat topped wave with 3rd harmonic. As S1 & S2 both are opened there is no closed path for the circulation of 3rd harmonic currents. So no compensating flux is produced for 3rd harmonic flux. Hence flux remains as flat topped wave.

=

0

19.

[Ans. D]

4 By applying KCL I1=25-10= 15A 20.

[Ans. C] Secondary current Secondary current referred to primary side (

)

As the core losses are neglected, magnetizing current (Im) will be in phase with flux ( ). Therefore Im lags the induced emf by Primary current + + . 4 6. 6

Volts

[Ans. A] i(t) = 10 sin (100π t) A induced emf on secondary E2 = M

1000V

To supply 10kVA Load I2 =

. t ch r) 1

D 15

B

[Ans. D] (slope of

25

400V

[Ans. C] With S2 closed, there is a closed path available for the 3rd harmonic currents within the phases. As the transformer is under no load condition only 3rd harmonic current will be flowing in the delta connected secondary.

V=M

17.

A

C

21.

[Ans. D] Transformer wise,

4 π

√

nd π cos(

4

cos(

πt)

4

sin .

π πt + /

πt )

So, turn ratio =

√ = 2√

So, Z =

+

( √ )

th

th

th


GATE QUESTION BANK

22.

[Ans. C] It is an air core transformer. So, there is no saturation effect.

23.

[Ans. B] We know

26.

Electrical Machines

[Ans. *] Range 545 to 555 urrent r ting of winding

uto tr nsformer r ting

√ π f as N and f are constant

27.

[Ans. A] Given data: , two winding transformer . } const nt f . f f ∴ ∝ f, ∝f f f . % ch nge f % f f % ch nge f ( . ) %

28.

[Ans. B]

nd lso i d √

⟹ ⟹i

√ i

⟹i

√

(

.

ore loss ∝ volume of core

) k

∴ .6

√

.

24.

[Ans. A] Hysteresis loss ∝

.

f

.

k

Hysteresis

k ( ) f

k f(

const nt)

f

.

k ( ) f

f

eddy

∝

f eddy loss ∝

f

k

f

k ( k (

8

+k (

)

+k (

k 86. 4 k Hysteresis k . ( eddy k ddy loss 4 .8 Hysteresis = 508 25.

)

8

[Ans. A] Thevenin volt ge

( ) sin t

Thevenin imped nce

( )

sin t ( )

const nt) (

+

)

. . core loss t f tf . k +k ( ) (k + k ) . . k +k ……… tf k +k ( ) (k + k ) k +k ……… k

8)

.

f

core loss t f

)

.

)

k ( ) f f

k f (

k

.

so

f

f

k

4Ω

t

th

Hz Hysteresis

th

th


GATE QUESTION BANK

eddy current

6 .

30.

(or) Alternative method v f. / f v f . / f v is const nt f f& f f+ f + f f From graph, at 0 Hz c f t Hz +

f

f

Hz f f

[Ans. B] Given data: - transformer 220/110V, 50Hz Self-inductance of primary 4 mH Self-inductance of Secondary mH Mutual inductance between primary and secondary M = 20 mH be the leakage & be the mutual inductance referred to primary side Equivalent diagram of inductance referred to primary

+ where

. t

Electrical Machines

.

(

)

w 6 . w tt (

29.

[Ans. *] Range 23 to 24 n econd ry side x x ( ) x 4 4k cos 4 .8 4 4 sin .4 6 .8 4 . 6 Energy supplied by reactance to make unity power factor 4 . 6 . / (

31.

4 . 6

) 4 . 6 4 4 4 . 6

.66

m)

( m

8 mH

m) 4 mH

[Ans. C] For same no load current f l Hl Number of turns of coil H Magnetic flux density l average length H∝ ∝ ∝ ∝ d dt So same voltage supply d d dt dt (πr

th

m)

4 m ( mH

th

)

(

πr 4

th

)


GATE QUESTION BANK

Electrical Machines

Induction Motors EE-2006 1. For a single phase capacitor start induction motor which of the following statements is valid? (A) The capacitor is used for power factor improvement (B) The direction of rotation can be changed by reversing the main winding terminals (C) The direction of rotation cannot be changed (D) The direction of rotation can be changed by interchanging the supply terminals 2.

3.

4.

The speed of a 4-pole induction motor is controlled by varying the supply frequency while maintaining the ratio of supply voltage to supply frequency (V/f) constant. At rated frequency of 50 Hz and rated voltage of 400 V its speed is 1440 rpm. Find the speed at 30 Hz, if the load torque is constant. (A) 882rpm (C) 840 rpm (B) 864 rpm (D) 828 rpm A 3 phase, 4 pole, 400 V, 50 Hz star connected induction motor has following circuit parameters r . Ω, r 0.5Ω, x = x ’ . Ω, x =35 Ω. The starting torque when the motor is started direct – on – line is (use approximate equivalent circuit model) (A) 63.6 Nm (C) 190.8 Nm (B) 74.3 Nm (D) 222.9 Nm

Blocked Rotor test : 400V 90 V 6A 15 A 1002 W 762 W Neglecting copper loss in no Load test and core loss in Blocked Rotor test, estimate motor’s full lo d efficiency. (A) 76 % (C) 82.4 % (B) 81 % (D) 85 % No Load test :

EE-2007 5. A three- phase squirrel cage induction motor has a starting torque of 150% and maximum torque of 300% with respect to rated torque at rated voltage and rated frequency. Neglect the stator resistance and rotational losses. The value of slip for maximum torque is (A) 13.48% (C) 18.92% (B) 16.42% (D) 26.79% EE-2008 6. A 230 V, 50 Hz 4-pole, single-Phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425-rpm. If the rotor resistance at stand still is 7.8Ω. Then the effective rotor resistance in the backward branch of the equivalent circuit will be (A) Ω (C) 8 Ω (B) 4 Ω (D) 6Ω 7.

A 400 V, 50 Hz, 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be (A) 23.06 kW (C) 25.01 kW (B) 24.11 kW (D) 26.21 kW

A 3 phase, 10 kW, 400 V, 4 pole, 50 Hz, star connected induction motor draws on full lo d. t’s no lo d nd blocked rotor test data are given below:

th

th

th


GATE QUESTION BANK

8.

A 400 V, 50 Hz, 4 pole, 1400 rpm, star connected squirrel cage induction motor has the Following parameters referred to the stator . Ω, = . Ω Neglect stator resistance and core and rotational losses of the motor The motor is controlled from a 3-phase voltage source inverter with constant V/f control. The stator line to line voltage (rms) and frequency to obtain the maximum torque at starting will be (A) 20.6 V, 2.7 Hz (B) 133.3 V, 16.7 Hz (C) 266.6 V, 33.3 Hz (D) 323.3V, 40.3 Hz Common Data for Questions 9 and 10 A 3-phase, 440 V 50Hz. 4-pole, slip ring induction motor is fed from the rotor side through an auto transformer and the stator is connected to variable resistance as shown in the figure.

Induction Motor

3 – phase, 50 Hz, Supply

Electrical Machines

9.

The speed of rotation of stator magnetic field with respect to rotor structure will be (A) 90 rpm in the direction of rotation, (B) 90 rpm in the opposite direction of rotation. (C) 1500 rpm in the direction of rotation (D) 1500 rpm in the opposite direction of rotation

10.

Neglecting all losses of both the machines, the dc generator power output and the current through resistance (Rex) will respectively be (A) 96 W, 3.10 A (B) 120 W, 3.46 A (C) 1504 W, 12.26 A (D) 1880 W, 13.71 A

EE-2009 11. A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor (solid curve) and of the load (dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B?

es

B

+ 220V

Torque

A

Auto Transformer

The motor is coupled to a 220 V, separately excited, dc generator feeding power to fixed power to fixed resistance of 10 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such that the motor runs at 1410 rpm and the following readings were recorded W1 = 1800 W, W2 =

0

(A) (B) (C) (D) 12.

th

1.0

sync

A is stable B is unstable A is unstable B is stable Both are stable Both are unstable

A 220V, 50Hz, single-phase induction motor has the following connection diagram and winding orientations shown. is the axis of the main stator winding th

th


GATE QUESTION BANK

(M1M2) and AA' is that of the auxiliary winding (A1A2). Directions of the winding axes indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated. When switch S is closed, the motor M1 r Ra

Ω

La = A1 A2

S 220V 50HZ

M2

H

. Ω m . /πH

A2 A2

Electrical Machines

EE-2011 14. A three-phase 440V, 6pole, 50 Hz, squirrel cage induction motor is running at a slip of 5%. The speed of stator magnetic field with respect to rotor magnetic field and speed of rotor with respect to stator magnetic field are (A) zero, 5 rpm (B) zero, 955 rpm (C) 1000 rpm, 5 rpm (D) 1000 rpm, 955 rpm EE-2012 15. The slip of an induction motor normally does not depend on (A) Rotor Speed (B) Synchronous speed (C) Shaft torque (D) Core-loss component

M

A Rotor ’

(A) (B) (C) (D)

rotates clockwise rotates anticlockwise does not rotate rotates momentarily and comes to a halt

EE-2010 13. A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by r , r , x , x and , respectively. The magnitude of starting current of the motor is given by (A) (B) (C) (D)

)

√( √

(

(

18.

)

(

The locked rotor current in a 3-phase, star connected 15 kW, 4-pole, 230 V, 50 Hz induction motor at rated conditions is 50 A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236 V, 57 Hz supply is (A) 58.5 A (C) 42.7 A (B) 45.0 A (D) 55.6 A

EE-2013 17. Leakage flux in an induction motor is (A) Flux that leaks through the machine (B) Flux that links both stator and rotor windings (C) Flux that links none of the windings (D) Flux that links the stator winding or the rotor winding but not both

) )

√( √

(

16.

) )

th

A 4-pole induction motor, supplied by a slightly unbalanced three-phase 50HZ source, is rotating at 1440 rpm. The electrical frequency in Hz of the induced negative sequence current in the rotor is (A) 100 (C) 52 (B) 98 (D) 48 th

th


GATE QUESTION BANK

EE-2014 19. An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 700 rpm. The frequency of the rotor current of the motor in Hz is __________ 20.

21.

22.

23.

A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1Ω and reactance of 0.92 Ω. Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is (A) 1.567 (C) 1.948 (B) 1.712 (D) 2.134

frequency operation of the motor is TRUE? (A) At low frequency, the stator flux increases from its rated value. (B) At low frequency, the stator flux decreases from its rated value. (C) At low frequency, the motor saturates. (D) At low frequency, the stator flux remains unchanged at its rated value 24.

A three-phase, 4-pole, self-excited induction generator is feeding power to a load at a frequency f . If the load is partially removed, the frequency becomes f . If the speed of the generator is maintained at 1500 rpm in both the cases, then (A) f , f Hz nd f f (B) f Hz nd f Hz (C) f , f Hz nd f f (D) f Hz nd f Hz A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MVAR is ______________

Electrical Machines

A three-phase slip-ring induction motor, provided with a commutator winding, is shown in the figure. The motor rotates in clockwise direction when the rotor windings are closed. -ph se c, f Hz

f rime over

lip ing nduction otor

f f

If the rotor winding is open circuited and the system is made to run at rotational speed f with the help of prime-mover in anti-clockwise direction, then the frequency of voltage across slip rings is f and frequency of voltage across commutator brushes is f . The values of f and f respectively are (A) f + f nd f (B) f f nd f (C) f f nd f + f (D) f + f nd f f 25.

In a constant V/f control of induction motor, the ratio V/f is maintained constant from 0 to base frequency, where V is the voltage applied to the motor at fundamental frequency f which of the following statements relating to low

th

A single phase induction motor is provided with capacitor and centrifugal switch in series with auxiliary winding. The switch is expected to operate at a speed of 0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speed characteristic of the motor is represented by

th

th


GATE QUESTION BANK

Electrical Machines

( )

Torque

.

x peed

( )

Torque

. peed

( )

Torque

.

peed

( )

Torque

. peed

th

th

th


GATE QUESTION BANK

Electrical Machines


[Ans. B]

4.

[Ans. B] The value of power loss (Cu loss) at full load

Switch

Rotor Auxiliary winding

=. / Im main winding

I

4.6 4

∴ Total losses = 1354.67 + 1002 = 2356.67

Start capacitor

Ia

6

∴ Efficiency =

.

= 81%

V

2.

Start capacitor is used to provide phase difference between nd . If supply terminals are interchanged, Ia and Im will flow in the opposite direction, so torque will act in the same direction. Therefore, the direction of rotation will remain same.

5.

[Ans. C]

6.

[Ans. D] T . T T T (

∝f

4

4 V

rpm 4 The slip (s) of rotor with respect to forward field

Now, T ∝ . / s ⟹

. / ×

Now, T

.

. 66 (

4

/

.

The slip of rotor with respect to backward field s . . Effective rotor resistance

) (

3.

s

T (given)

. 4

= 26.79%.

[Ans. A] Rotor resistance at stand still R = 7.8 Synchronous speed f

= constant (given) ⟹

)

∴4 ( ) + ( ) 4 + ∴ Correct answer is,

. 66) = 840.6 rpm

[Ans. A] (

= 1500 rpm T

(

) (

/√ )

( . )

s) .8 .

( . )

.

7.

[Ans. C] Input power in stator =√ 4 .8 . kW Air gap power = 27.71 – 1.5 – 1.2 = 25.01 kW

= 63.58 Nm

th

th

th


GATE QUESTION BANK

8.

[Ans. B]

12.

For max. torque slip = For starting torque + πf + . πf (frequency at maximum f

(

torque)

) .

[Ans. B] f= 50 Hz Impedance of main winding Z r + j πf . . +j π π Z . +j Impedance of auxiliary winding Z r + j πf +j π

.

∴ f

Electrical Machines

. /

π Z +j Current through main winding

. /

= Z

. +j 8 .4 Current through auxiliary winding

= 16.7 Hz In const V/f control method ∴ ∴

/f

8

8

f 8 = 16.7 × 8 = 133.3 volts

Z

+j . 8 . 4 Taking Vs as the reference V

9.

[Ans. A] = 1500 rpm In the direction of rotation of rotor, therefore, speed of the stator field with respect to rotor 1500 1410 = 90 rpm

Ia Im

10.

[Ans. C] Slip =

Im leads Ia the fields created by the two current also have same difference thereby constituting an unbalanced field system. The result is the production of the starting torque. Space – orientation of the fied

= 0.06

Total power input to induction motor within = 1800 – 200 = 1600 W Power output of induction motor ( s) 6 ( . 6) = 1504 W

11.

ex

4

√

4

Field due to Im

∴

Field due to Ia

. 6

[Ans. A] t point if speed ↑ lo d torque lso ↑ nd if speed ↓ lo d torque ↓ hence is st ble but t if speed ↑ lo d torque ↓ s vice versa an were b is unstable.

The motor rotates in the direction of leading phase to lagging phase. In this case, the motor rotates anti – clockwise.

th

th

th


GATE QUESTION BANK

13.

[Ans. A] or the circuit.

Electrical Machines

17.

[Ans. D]

18.

[Ans. B] f

So it can be neglected from v

√(r + r ) + (x + x )

(

)f

f

r

. 4 (

f

. 4)

8Hz

r

19.

V

14.

[Ans. *]Not matching with IIT keys Stator and rotor magnetic fields rotates at same speed. So difference in speed is zero. Speed of stator magnetic field is Ns (synchronous speed) f rpm p 6 Rotor speed = ( s) rpm So, speed of rotor with respect to stator magnetic field

sf 20.

[Ans. C] T T s

[Ans. D] Slip , S =

16.

.

+

slip t which m x torque ppe rs

x s s . 8 full lo d slip

Where, = synchronous speed – rotor speed As the shaft torque depends upon rotor speed therefore the slip also depends on shaft torque. And core-losses are independent of slip

sf

s

rpm. 15.

[Ans. *] Range 3.2 to 3.5 Frequency of rotor current of motor f supply frequency s slip of motor f rpm 8

( .

T o T 21.

[Ans. B] At standstill , the rotor current is √ + As losses are zero

.

. )

+.

. .

/

.

8

[Ans. C] Given data: , 4 pole self excited induction generator, feeding power at frequency f Condition 1:In induction generator speed of generator 1500rpm ∴ rpm i. e. , rpm f

if ut

rpm , f rpm

f

Hz

Hz

Condition 2:If load is partially removed and frequency f speed of generator is constant therefore rpm, but load

∴ 4

th

th

th


GATE QUESTION BANK

removed means speed of the RMF is increased compared to previous condition ∴f Hz but f f ∴ f ,f Hz, f f

25.

Electrical Machines

[Ans. C] capacitor start induction motor

entrifug l switch

in winding A.C supply

22.

[Ans. *] Range 6.97 to 7.03 At 0.6 lagging power factor cos .6



.6 sin

.8 .6 At 0.8 lagging power factor cos (re l power rem in s me) .8



.8

 .6

.8

ue to m in + uxili ry + switch

.8 .6

.6

in winding only

.8

Torque

6 23. 24.

The currents induced in both main, auxiliary winding with 90 phase displacement. So that as per induction principle R.M.F will produce and then M/c will rotates with some starting torque (T ) Centrifugal switch is designed here when 70% of reached it will open i.e., auxiliary winding is removed electrically and M/c will rotates as per doubled field revolving theory. Torque slip characteristics as shown in figure below

[Ans. B] . peed

[Ans. A] , slip ring induction motor, provided with commutator winding motor rotates in clockwise direction when the rotor windings are closed. i.e., in motor operation Now rotor winding is open circuited and the system is made to run at rotational speed f in anticlockwise direction. In this condition operates as induction generator so rotor rotates a head of stator R.M.F and therefore frequency across slip rings is f+f f frequency of voltages across commutator brushes is f f



uxili ry

Due to malfunctioning centrifugal switch not opened, so that auxiliary winding also present even motor reaches greater than 70% so that torque slip characteristics will change as shown in figure below

Torque

. s peed

th

th

th


GATE QUESTION BANK

Electrical Machines

D.C. Machine EE-2006 1. In a DC machine, which of the following statements is true? (A) Compensating winding is used for neutralizing armature reaction while inter pole winding is used for producing residual flux (B) Compensating winding is used for neutralizing armature reaction while inter pole winding is used for improving commutation (C) Compensating winding is used for improving commutation while inter pole winding is used for neutralizing armature reaction (D) Compensating winding is used for improving commutation while inter pole winding is used for producing residual flux 2.

A 220 V DC machine supplies 20 A at 200 V as a generator. The armature resistance is 0.2 ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10% then ratio of motor speed to generator speed is (A) 0.87 (C) 0.96 (B) 0.95 (D) 1.06

80rad/s. The rm ture esist nce is . Ω nd the field winding resist nce is 8 Ω. 4.

The net voltage across the armature resistance at the time of plugging will be (A) 6V (C) 240 V (B) 234V (D) 474 V

5.

The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is (A) . Ω (C) . Ω (B) . Ω (D) . Ω

EE-2009 6. Figure shows the extended view of a 2 pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the inter polar axis. If the brushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform will resemble N

’

B

1

EE-2007 3. The dc motor, which can provide zero speed regulation at full load without any controller is (A) series (B) shunt (C) cumulative compound (D) differential compound

S

’ +

A +

2

3

4

’ 5 5 ot tion t speed

’

’ 4’

’

r d/sec

(A) V ’ ’

t . π

(B)

V

EE-2008 Statement for Linked Answer Questions 4 and 5 A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of

.4π

.6π

.8π

π

’ ’

t . π

th

th

.4π

th

.6π

.8π

π


GATE QUESTION BANK

(C) V

Electrical Machines

The rotational losses and armature reaction are neglected.

’ ’

Speed (rpm) 1500 t . π

.4π

.6π

.8π

1400

π

(D) V

’ ’

t . π

.4π

.6π

.8π

EE-2010 7. A separately excited dc machine is coupled to a 50Hz, three-phase, 4-pole induction machine as shown in the figure. The dc machine is energized first and the machines rotate at 1600 rpm. Subsequently the induction machine is also connected to a 50Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state, DC machine

Induction machine 4 pole, 50 Hz

0

π

50 Hz, balanced three-phase supply

(A) both machines act as generators (B) the dc machine acts as a generator, and induction machine acts as a motor (C) the dc machine acts as a motor, and the induction machine acts a generator (D) both machines act as motors Common Data for Questions 8 and 9 A separately excited DC motor runs at 1500 rpm under no-load with 200V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers at torque of 5 Nm, is 1400 rpm as shown in the figure.

5

torque (Nm)

8.

The armature resistance of the motor is. (A) Ω (C) 4.4Ω (B) .4Ω (D) . Ω

9.

For the motor to deliver a torque of 2.5Nm at 1400 rpm, the armature voltage to be applied is (A) 125.5V (C) 200V (B) 193.3V (D) 241.7V

EE-2011 10. A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1.0 Ω and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extra resistance to be put in the armature circuit to maintain the same speed and torque will be (A) . Ω (C) . Ω (B) . Ω (D) 8. Ω 11.

A 4-point starter is used to start and control the speed of a (A) dc shunt motor with armature resistance control (B) dc shunt motor with field weakening control (C) dc series motor (D) dc compound motor

EE-2012 12. A 220 V, 15 kW, 1000 rpm shunt motor with armature resistance of 0.25 Ω, has a rated line current of 68 A and a rated field th

th

th


GATE QUESTION BANK

current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is (A) 18.18 % increase (B) 18.18 % decrease (C) 36.36 % increase (D) 36.36 % decrease

15.

A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5Ω. and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is_________

16.

A 15 kW, 230 V dc shunt motor has armature circuit resistance of 0.4 Ω and field circuit resistance of 230 Ω. At no load and rated voltage, the motor runs at 1400 rpm and the line current drawn by the motor is 5 A. At full load, the motor draws a line current of 70 A. Neglect armature reaction. The full load speed of the motor in rpm is____________

EE-2014 13. A 250 V dc shunt machine has armature circuit resistance of 0.6 Ω and field circuit resistance of 125 Ω. The machine is connected to 250 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both the cases is 50 A. The ratio of the speed as a generator to the speed as a motor is ___ 14.

Electrical Machines

The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The armature resistance drop and the brush drop are neglected. The field current is kept constant at rated value. The torque of the motor in Nm for an armature current of 8 A is __________

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GATE QUESTION BANK

Electrical Machines


2.

[Ans. B] Compensating winding is placed in slots cut out in pole faces such that the axis of this winding coincides with the brush axis. The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization. To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In order that this injection is restricted to commutating coils, narrow interpolar are provided in the interpolar region. [Ans. A] For generator E=V+ + For motor V=E+ E = 200 – 20 × 0.2 = 196

4.

[Ans. D] 240 V, shunt motor 15A, N = 80 rad/sec = 0.5 8 Ω 4 . = 234 plugging + 4 + 4 4 4

5.

[Ans. A] Here,

.

+ ∴ 6.

.6 = 31.1

[Ans. A] When brushes are shifted in the direction of motor rotation then rise time less than fall time.

e ding tip

Tr iling tip

Tr iling tip

e ding tip

. = 204

.

∴ 3.

.

= 0.87

[Ans. D] Speed – current characteristics of DC motors Speed

orce

DC machines is acting as DC motor When the brush is shifted in the direction of rotation in the DC motor, field gets magnetized. Due to armature reaction, leading tip of N-pole and S-pole demagnetized. But increase in flux density. Due to magnetized effect of armature reaction waveform of will be as given option (a)

Differential compound n0

Shunt Cumulative comp. Series Ia (Ia)full load

It can seen from the characteristics, speed regulation can be zero at full load in the case of differential component dc motor. th

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GATE QUESTION BANK

7.

[Ans. C]

N∝

Synchronous speed

[Ans. B] At no load

, = 200 V

.

⟹

12.

[Ans. D]

86.6

.

6 .8

.8 .

+

[Ans. B] 4-point starter is used to control the speed of shunt motor in field weakening region. i.e.; above rated speeds. In field weakening region field current will reduce in 3 – point starter holding coil unable to hold the plugges in ON position.

4

.

)

11. rpm,

86.6

. ]⟹ ( + ) . Ω

. .

86.6 V

. .

(

[

Assuming ∝ , at N = 14010 rpm 4

T

⟹

.

=1500 rpm So, slip S < 0 Hence, dc machine acts as a motor and Induction machine acts as a generator. 8.

Electrical Machines

.

0

Ω

. .

6 .8

1

.

.6 64 9.

[Ans. B] For N = 1400 rpm, 86.6 equired ‘ ’ + 86.6 + ( . ) Where is function of torque, To Develop 5 N-m it requires 3.925 A . . ( )

%decrease= *

10.

)( .

+ 6. 6%decre se

13.

. . 6 86.6 + ( . 6 = 193.3 V

x

)

[Ans. *] Range 1.22 to 1.32 Given data: Dc shunt machine operated as generator and motor separately .6Ω Ω

[Ans. A] Find: ratio of speed as a generator to speed as motor

Now flux is decreased by 10%, so . Torque is constant so ⟹

.

n gener tor + +

.

+ +

th

th

.6 th


GATE QUESTION BANK

Electrical Machines

8 . In motor operation:

. + . k k

48

. .

48 .6 . speed s gener tor peed s motor

∴

+ f

k

b

4 At no load .4 +

k

4 Field current is kept constant at rated value so k will be same at I = 8A ki (8

(8)

( t r ted volt ge) 4 4 .4 + 8.4 At rated load r + ( ) .4 + .4 k .4 k 4 8.4 4 .6 rpm

4 6 / π) in m

4 .

15.

iled winding

+

[Ans. *] Range 12.45 to 12.65 o lo d

T

[Ans. *] Range 1239 to 1242

n shunt ⁄c] 8 . . .

[

14.

16.

88 .4 rpm

m

[Ans. *] Range 879 to 881 No load r + . +

,

k -

k under loaded condition

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GATE QUESTION BANK

Electrical Machines

Synchronous Machine EE-2006 1. A synchronous generator is feeding a zero power factor (lagging) load at rated current. The armature reaction is (A) magnetizing (B) demagnetizing (C) cross – magnetizing (D) ineffective 2.

A 3 phase, 400 V, 5 kW, star connected synchronous motor having an internal reactance of 10Ω is operating at 50% load, unity pf. Now, the excitation is increased by 1%. What will be the new load in percent, if the power factor is to be kept same? Neglect all losses and consider linear magnetic circuit. (A) 67.9% (C) 51% (B) 56.9% (D) 50% Common Data Question for 3, 4 and 5 A 4 pole, 50 Hz, synchronous generator has 48 slots in which a double layer winding is used. Each coil has 10 turns and is short pitched by an angle to 6 electrical. The fundamental flux per pole is 0.025 Wb.

3.

4.

5.

The line-to-line induced emf (in volts), for a three phase star connection is approximately (A) 808 (C) 1400 (B) 888 (D) 1538 The line-to-line induced emf (in volts), for a three phase connection is approximately (A) 1143 (C) 1617 (B) 1332 (D) 1791

EE-2007 6. A three-phase synchronous motor connected to ac main is running at full load and unity power factor. If its shaft load is reduced by half, with field current held constant, its new power factor will be (A) unity (B) leading (C) lagging (D) dependent on machine parameters 7.

A 100kVA. 415 V (line), Star-connected synchronous machine generates rated open circuit voltage of 415 V at field current of 15A. The short circuit armature current at a field current of 10 A is equal to the rated armature current. The per unit saturated synchronous reactance is (A) 1.731 (C) 0.666 (B) 1.5 (D) 0.577

EE-2008 8. Distributed winding and short chording employed in AC machines will result in (A) Increase in emf and reduction in harmonics. (B) Increase in emf and increase in harmonics. (C) Increase in both emf and harmonics. (D) Reduction in both emf and harmonics. Statement for Linked Answer Questions 9 and 10 A synchronous motor is connected to an infinite bus at 1.0pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is 1.0pu and resistance is negligible. The excitation voltage (E) and load angle ( ) will respectively be (A) 0.8 pu and 36.86 lag (B) 0.8pu and 36.86o lead (C) 1.17 pu and 30.96° lead (D) 1.17 pu and 30.96° lag

9.

The fifth harmonic component of phase emf (in volts), for a three phase star connection is (A) 0 (C) 281 (B) 269 (D) 808 th

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GATE QUESTION BANK

10.

Keeping the excitation voltage same, the load on the motor is increased such that the motor current increase by 20%. The operating power will become (A) 0.995 lagging (C) 0.791 lagging (B) 0.995 leading (D) 0.848 leading

EE-2009 11. A field excitation of 20 A in a certain alternator results in an armature current of 400A in short circuit and a terminal voltage of 2000V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200A is (A) 1V (C) 100V (B) 10V (D) 1000V EE-2011 12. The direct axis and quadrature axis reactances of a salient pole alternator are 1.2 p.u and 1.0 p.u respectively. The armature resistance is negligible. If this alternator is delivering rated kVA at upf and at rated voltage then its power angle is (A) 30° (C) 60° (B) 45° (D) 90° 13.

A three-phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load at normal excitation. The field excitation of the motor is first reduced to zero and then increased in the reverse direction gradually. Then the armature current (A) Increase continuously (B) First increases and then decreases steeply (C) First decreases and then increases steeply (D) Remains constant

Electrical Machines

performed on it with 90% of the rated current flowing in its both LV and HV windings, the measured loss is 81 W. The transformer has maximum efficiency when operated at (A) 50.0% of the rated current. (B) 64.0% of the rated current. (C) 80.0% of the rated current. (D) 88.8% of the rated current. 15.

The angle in the swing equation of a synchronous generator is the (A) Angle between stator voltage and current (B) Angular displacement of the rotor with respect to the stator (C) Angular displacement of the stator mmf with respect to a synchronously rotating axis. (D) Angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis.

EE-2014 16. A star connected 400 V, 50 Hz, 4 pole synchronous machine gave the following open circuit and short circuit test results: Open circuit test: = 400 V (rms, line-to-line) at field current, = 2.3 A Short circuit test: = 10 A (rms, phase) at field current, = 1.5 A The value of per phase synchronous impedance in Ω at rated voltage is____________ 17.

A three phase synchronous generator is to be connected to the infinite bus. The lamps are connected as shown in the figure for the synchronization. The phase sequence of bus voltage is R-Y-B and that of incoming generator voltage is R'-Y'-B'.

EE-2013 14. A single-phase transformer has no-load loss of 64 W, as obtained from an opencircuit test. When a short-circuit test is th

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GATE QUESTION BANK Y

~ nfinite us

19.

In a synchronous machine, hunting is predominantly damped by (A) mechanical losses in the rotor (B) iron losses in the rotor (C) copper losses in the stator (D) copper losses in the rotor

20.

A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is ____

~

Y

~ ncoming ener tor

It was found that the lamps are becoming dark in the sequence - - . It means that the phase sequence of incoming generator is (A) opposite to infinite bus and its frequency is more than infinite bus (B) opposite to infinite bus but its frequency is less than infinite bus (C) same as infinite bus and its frequency is more than infinite bus (D) same as infinite bus and its frequency is less than infinite bus 18.

Electrical Machines

A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generated voltage is . If the coils are reconnected to realize three-phase star-connected winding, the generated phase voltage is . Assuming full pitch, single-layer winding, the ratio / is (A) /√ (C) √ (B) / (D)

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GATE QUESTION BANK

Electrical Machines


[Ans. B] Xs

Ef

cos α / cos 8 = 0.951 n double layer wdg. no. of coils = no. of slots

Ia

~

Vt

jIa

no. of turns/ph =

Vt

. ∴

Ia

Ef > Vt, the generator is overexcited. Therefore armature – reaction is demagnetizing in nature i.e. ar opposes

4.

8 8 √ = 1400V

or Synchronous Motor, ( cos ) + ( sin ∴ E = √.

√

6

Truns/ph

7.21

√

/ +(

4

Slot angle = sin(

)

)

.

6 sin . /

.6)

= 2333.729 If these is (0.01) increase in becomes, . . = 236

cos ( , then

6

)

.

4.44f

T

4.44

.

. .

=√

= 1143 v

=√ ∴

)

6

Slots/pole/ph

[Ans. A]

∴

(

T

[Ans. C ] For two phase scott connection

f.

2.

4.44 f 4.44 . . = 808 V

Is

6

6

(4

j

/√ ) = 48.93

4 v

4.8

∴ % load =

.

= 67.86%

.

4 v

3.

[Ans. C] 4 pole, 50 Hz, no. of slot = S = 48 double layer winding, 1 coil contains 10 turns, α(short pitched) 6 , /pole = 0.025 mwb. 4.44 f T slot /pole/ph = = 12

slot angle =

=

(

/ ) (

/ )

5.

[Ans. A] Pitch factor component

=4

slot/pole =

4 √ = 1616.45 V

√

(

)

cos (

due 6

to )

5th

harmonic

cos

[as pitch factor is zero, induced emf due to 5th harmonic component is zero]

= 0.957 th

th

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GATE QUESTION BANK

6.

[Ans. B]

8.

[Ans. D] Induced emf 4.44k k fT As KP & kd v lues re less th n ‘ ’ for distributed & short pitched winding, the induced emf reduces. To eliminate higher order harmonics we use distributed winding and to eliminate the dominant harmonics i.e., 5th & 7th we use pitched winding.

9.

[Ans. D] V = 1 ∠0 pu .6∠ z +j +j ∠ pu ∠ ∠ .6∠ ∠ E∠ . 66 ∠ . 6 pu (It is lagging) ∴ excitation voltage = 1.17 pu ∴ lo d ngle ( ) . 6

10.

[Ans. A] Let power factor = cos l gging .6 . pu ∠ j ∠ (cos + j sin ) (cos ∠ j j sin ) (cos j . j sin ) ( . sin ) j . cos ( . sin ) + ( . cos ) + . sin .44 sin + . cos . + . .44 sin sin . 8 . 6 cos . l gging

11.

[Ans. D]

Xs

Ia

~

Vt

Electrical Machines

Ef

Neglecting armature resistance Ra Vt=Ef +jIa Xs Ia2 Vt

Ia1

jIa2Xs jIa1Xs Ef

Reduced load Ef at shaft load at shaft

It can be seen from the phasor diagram that, when load is reduced at shaft keeping field current constant the new power factor becomes leading. 7.

[Ans. C] Saturation of field poles occur for a field current equal to for which rated open circuited voltage is obtained. ∴ for is ( ) for ∴

(

(

)

√

for ∴Z

(

)

(

p.u Z

)

Z

)

. 4 8.68 4 . 88 ohm 8.68 . . 88 6.66 .

(

)

.666

2nd Method Z

( (

Z

) ⁄ph

( . )

)

(4

(Z ⁄ph) .6 8 (4

⁄√ )

Internal resistance =

⁄√ ) . 4 8.68 .6 8 . .666

Ω

Internal voltage drop = 5×200 = 1000V 12.

[Ans. B] (

T n

) (

p. u angle = 0° x . p. u x T n ⟹ th

th

)

p. u

Power factor

p. u r 4 th


GATE QUESTION BANK

13.

[Ans. B] As field current reduces the flux will start reducing, to keep this flux constant the armature draws high current from bus. At zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus, if we increase field current in reverse direction to keep flux constant motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slips one pole pitch and align to opposite pole. The instant aligns to opposite pole the flux will be very high to reduce this flux current drops steeply to synchronous motor value. Here torque is reluctance + synchronous motor torque.

14.

[Ans. C] Cu.Loss at 90% load = 81W Cu load at any X fractional load full lo d cu. loss full lo d cu. loss

x

√

17.

[Ans. A] In Synchronization [In figure lights are connected in correct position i.e., y y, ] 1. Observe the lights which are connected across switches, they should beat, first get brighter and then dim as the phasor for generator nd bus respectively shift. f ‘ ’ lights beats concurrently, the phase sequence is correct. Else if lights beat out of phase, one pair of phases should be reversed 2. Adjust now the prime mover to slowly increase/decrease the speed of incoming generator. One should observe a slow beat of the light brightness 3. When the lamps beats slowly, the switches should be closed when the lights are extinguished In given data lamps are becoming dark in the sequence it means that phase sequence of incoming generator is opposite to infinite bus and its frequency is more than infinite bus.

18.

[Ans. D] Given data: 20 pole alternator Total number of turns (T) = 8 6 or T ph se spre d 8 (mv) T or ph se spre d

( . )

For maximum efficiency ) 64 x ( .8

8 % of r ted lo d 15.

[Ans. D]

16.

[Ans. *] Range 14.5 to 15.5 ( t . ) 4 ( ) 4 h se √ . 4 ( t . ) . √ 4 . in Ω(per ph se) . √ . 6Ω

Electrical Machines

6 winding gener ted volt ge

n

winding gener ted ph se volt ge We know that T where Generated voltage in phase manner k Distribution factor T Number of turns per phase th

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GATE QUESTION BANK

sin sin

19.

(

)

(

)

sin 8 / sin 6 /

6 T . 8 T/

20.

/

Electrical Machines

[Ans. *] Range 32.4 to 34.0 (pu) (pu) (pu) cos . ∠ ( ) . power f ctory unity ∠ . ∠ . ∠ .8∠ ∠ . ∠ . ∠ ∠ . + . j . t n . .468

[Ans. D] In a synchronous machine, hunting is predominantly damped by damper winding in other words when hunting occurs in a system as per induction principle a currents is induced in damper winding those will give either induction generator torque or motor torque. Which is in phase/opposite phase to rotation as per options by copper losses in rotor hunting is damped because induced currents will give losses also. In case of salient pole synchronous machine. In case of cylindrical generator no damper winding present rotor core itself act as good damper bar and induced current called eddy currents.

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GATE QUESTION BANK

Electrical Machines

Principles of Electro Mechanical Energy Conversion EE-2007 1. The electromagnetic torque Te of a drive, and its connected load torque TL are as shown below. Out of the operating points A, B, C and D, the stable ones are (A)

Te

TL

T A

Speed (B)

Te

T

TL B

Speed (C)

T

Te

C TL Speed

(D)

Te T D TL Speed

(A) A, C, D (B) B, C

(C) A, D (D) B, C, D

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GATE QUESTION BANK

Electrical Machines


[Ans. A] Accelerating torque = T T Te TL

T

So, rotor accelerates and speed increases. It speed increases due to some disturbances T T T T T o, rotor decelerates and speed decreases o, point A, C, D are stable.

T

A

T T

Speed T

T

C

B

Te

T

Speed

At point B, T T T T T If speed decreases, due to some disturbance T T T T T So rotor decelerates, and rotor speed keeps on decreasing. If speed increases. T T T T T Rotor accelerates, and rotor speed keeps on increasing. So, point is unstable

Speed T

Te D T Speed

AT points A, C, D T T T T T It due to some disturbance speed decreases T T T T T

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GATE QUESTION BANK

Electrical Machines

Special Machines EE-2007 1. A three-phase, three-stack, variable reluctance step motor has 20 poles on each rotor and stator stack , The step angle of this step motor is (A) (C) (B) 6 (D) 8 EE-2008 2. In a stepper motor, the detent torque means (A) minimum of the static torque with the phase winding excited. (B) maximum of the static torque with the phase winding excited. (C) minimum of the static torque with the phase winding unexcited. (D) maximum of the static torque with the phase winding unexcited


[Ans. B] Step angle =

2.

6

[Ans. D] Detent Torque (or) Restraining Torque is the maximum Load Torque that can be kept on motor shaft in an unexcited motor without causing continuous rotation.

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GQB_EE.pdf

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